I’d like to see integers, positive or negative, in binary.
Rather like this question, but for JavaScript.
function dec2bin(dec) {
return (dec >>> 0).toString(2);
}
console.log(dec2bin(1)); // 1
console.log(dec2bin(-1)); // 11111111111111111111111111111111
console.log(dec2bin(256)); // 100000000
console.log(dec2bin(-256)); // 11111111111111111111111100000000
You can use Number.toString(2) function, but it has some problems when representing negative numbers. For example, (-1).toString(2) output is "-1".
To fix this issue, you can use the unsigned right shift bitwise operator (>>>) to coerce your number to an unsigned integer.
If you run (-1 >>> 0).toString(2) you will shift your number 0 bits to the right, which doesn't change the number itself but it will be represented as an unsigned integer. The code above will output "11111111111111111111111111111111" correctly.
This question has further explanation.
-3 >>> 0 (right logical shift) coerces its arguments to unsigned integers, which is why you get the 32-bit two's complement representation of -3.
Try
num.toString(2);
The 2 is the radix and can be any base between 2 and 36
source here
UPDATE:
This will only work for positive numbers, Javascript represents negative binary integers in two's-complement notation. I made this little function which should do the trick, I haven't tested it out properly:
function dec2Bin(dec)
{
if(dec >= 0) {
return dec.toString(2);
}
else {
/* Here you could represent the number in 2s compliment but this is not what
JS uses as its not sure how many bits are in your number range. There are
some suggestions https://stackoverflow.com/questions/10936600/javascript-decimal-to-binary-64-bit
*/
return (~dec).toString(2);
}
}
I had some help from here
A simple way is just...
Number(42).toString(2);
// "101010"
The binary in 'convert to binary' can refer to three main things. The positional number system, the binary representation in memory or 32bit bitstrings. (for 64bit bitstrings see Patrick Roberts' answer)
1. Number System
(123456).toString(2) will convert numbers to the base 2 positional numeral system. In this system negative numbers are written with minus signs just like in decimal.
2. Internal Representation
The internal representation of numbers is 64 bit floating point and some limitations are discussed in this answer. There is no easy way to create a bit-string representation of this in javascript nor access specific bits.
3. Masks & Bitwise Operators
MDN has a good overview of how bitwise operators work. Importantly:
Bitwise operators treat their operands as a sequence of 32 bits (zeros and ones)
Before operations are applied the 64 bit floating points numbers are cast to 32 bit signed integers. After they are converted back.
Here is the MDN example code for converting numbers into 32-bit strings.
function createBinaryString (nMask) {
// nMask must be between -2147483648 and 2147483647
for (var nFlag = 0, nShifted = nMask, sMask = ""; nFlag < 32;
nFlag++, sMask += String(nShifted >>> 31), nShifted <<= 1);
return sMask;
}
createBinaryString(0) //-> "00000000000000000000000000000000"
createBinaryString(123) //-> "00000000000000000000000001111011"
createBinaryString(-1) //-> "11111111111111111111111111111111"
createBinaryString(-1123456) //-> "11111111111011101101101110000000"
createBinaryString(0x7fffffff) //-> "01111111111111111111111111111111"
This answer attempts to address inputs with an absolute value in the range of 214748364810 (231) – 900719925474099110 (253-1).
In JavaScript, numbers are stored in 64-bit floating point representation, but bitwise operations coerce them to 32-bit integers in two's complement format, so any approach which uses bitwise operations restricts the range of output to -214748364810 (-231) – 214748364710 (231-1).
However, if bitwise operations are avoided and the 64-bit floating point representation is preserved by using only mathematical operations, we can reliably convert any safe integer to 64-bit two's complement binary notation by sign-extending the 53-bit twosComplement:
function toBinary (value) {
if (!Number.isSafeInteger(value)) {
throw new TypeError('value must be a safe integer');
}
const negative = value < 0;
const twosComplement = negative ? Number.MAX_SAFE_INTEGER + value + 1 : value;
const signExtend = negative ? '1' : '0';
return twosComplement.toString(2).padStart(53, '0').padStart(64, signExtend);
}
function format (value) {
console.log(value.toString().padStart(64));
console.log(value.toString(2).padStart(64));
console.log(toBinary(value));
}
format(8);
format(-8);
format(2**33-1);
format(-(2**33-1));
format(2**53-1);
format(-(2**53-1));
format(2**52);
format(-(2**52));
format(2**52+1);
format(-(2**52+1));
.as-console-wrapper{max-height:100%!important}
For older browsers, polyfills exist for the following functions and values:
Number.isSafeInteger()
Number.isInteger()
Number.MAX_SAFE_INTEGER
String.prototype.padStart()
As an added bonus, you can support any radix (2–36) if you perform the two's complement conversion for negative numbers in ⌈64 / log2(radix)⌉ digits by using BigInt:
function toRadix (value, radix) {
if (!Number.isSafeInteger(value)) {
throw new TypeError('value must be a safe integer');
}
const digits = Math.ceil(64 / Math.log2(radix));
const twosComplement = value < 0
? BigInt(radix) ** BigInt(digits) + BigInt(value)
: value;
return twosComplement.toString(radix).padStart(digits, '0');
}
console.log(toRadix(0xcba9876543210, 2));
console.log(toRadix(-0xcba9876543210, 2));
console.log(toRadix(0xcba9876543210, 16));
console.log(toRadix(-0xcba9876543210, 16));
console.log(toRadix(0x1032547698bac, 2));
console.log(toRadix(-0x1032547698bac, 2));
console.log(toRadix(0x1032547698bac, 16));
console.log(toRadix(-0x1032547698bac, 16));
.as-console-wrapper{max-height:100%!important}
If you are interested in my old answer that used an ArrayBuffer to create a union between a Float64Array and a Uint16Array, please refer to this answer's revision history.
A solution i'd go with that's fine for 32-bits, is the code the end of this answer, which is from developer.mozilla.org(MDN), but with some lines added for A)formatting and B)checking that the number is in range.
Some suggested x.toString(2) which doesn't work for negatives, it just sticks a minus sign in there for them, which is no good.
Fernando mentioned a simple solution of (x>>>0).toString(2); which is fine for negatives, but has a slight issue when x is positive. It has the output starting with 1, which for positive numbers isn't proper 2s complement.
Anybody that doesn't understand the fact of positive numbers starting with 0 and negative numbers with 1, in 2s complement, could check this SO QnA on 2s complement. What is “2's Complement”?
A solution could involve prepending a 0 for positive numbers, which I did in an earlier revision of this answer. And one could accept sometimes having a 33bit number, or one could make sure that the number to convert is within range -(2^31)<=x<2^31-1. So the number is always 32bits. But rather than do that, you can go with this solution on mozilla.org
Patrick's answer and code is long and apparently works for 64-bit, but had a bug that a commenter found, and the commenter fixed patrick's bug, but patrick has some "magic number" in his code that he didn't comment about and has forgotten about and patrick no longer fully understands his own code / why it works.
Annan had some incorrect and unclear terminology but mentioned a solution by developer.mozilla.org
Note- the old link https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators now redirects elsewhere and doesn't have that content but the proper old link , which comes up when archive.org retrieves pages!, is available here https://web.archive.org/web/20150315015832/https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators
The solution there works for 32-bit numbers.
The code is pretty compact, a function of three lines.
But I have added a regex to format the output in groups of 8 bits. Based on How to format a number with commas as thousands separators? (I just amended it from grouping it in 3s right to left and adding commas, to grouping in 8s right to left, and adding spaces)
And, while mozilla made a comment about the size of nMask(the number fed in)..that it has to be in range, they didn't test for or throw an error when the number is out of range, so i've added that.
I'm not sure why they named their parameter 'nMask' but i'll leave that as is.
https://web.archive.org/web/20150315015832/https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_Operators
function createBinaryString(nMask) {
// nMask must be between -2147483648 and 2147483647
if (nMask > 2**31-1)
throw "number too large. number shouldn't be > 2**31-1"; //added
if (nMask < -1*(2**31))
throw "number too far negative, number shouldn't be < -(2**31)" //added
for (var nFlag = 0, nShifted = nMask, sMask = ''; nFlag < 32;
nFlag++, sMask += String(nShifted >>> 31), nShifted <<= 1);
sMask=sMask.replace(/\B(?=(.{8})+(?!.))/g, " ") // added
return sMask;
}
console.log(createBinaryString(-1)) // "11111111 11111111 11111111 11111111"
console.log(createBinaryString(1024)) // "00000000 00000000 00000100 00000000"
console.log(createBinaryString(-2)) // "11111111 11111111 11111111 11111110"
console.log(createBinaryString(-1024)) // "11111111 11111111 11111100 00000000"
//added further console.log example
console.log(createBinaryString(2**31 -1)) //"01111111 11111111 11111111 11111111"
You can write your own function that returns an array of bits.
Example how to convert number to bits
Divisor| Dividend| bits/remainder
2 | 9 | 1
2 | 4 | 0
2 | 2 | 0
~ | 1 |~
example of above line: 2 * 4 = 8 and remainder is 1
so 9 = 1 0 0 1
function numToBit(num){
var number = num
var result = []
while(number >= 1 ){
result.unshift(Math.floor(number%2))
number = number/2
}
return result
}
Read remainders from bottom to top. Digit 1 in the middle to top.
This is how I manage to handle it:
const decbin = nbr => {
if(nbr < 0){
nbr = 0xFFFFFFFF + nbr + 1
}
return parseInt(nbr, 10).toString(2)
};
got it from this link: https://locutus.io/php/math/decbin/
we can also calculate the binary for positive or negative numbers as below:
function toBinary(n){
let binary = "";
if (n < 0) {
n = n >>> 0;
}
while(Math.ceil(n/2) > 0){
binary = n%2 + binary;
n = Math.floor(n/2);
}
return binary;
}
console.log(toBinary(7));
console.log(toBinary(-7));
You could use a recursive solution:
function intToBinary(number, res = "") {
if (number < 1)
if (res === "") return "0"
else
return res
else return intToBinary(Math.floor(number / 2), number % 2 + res)
}
console.log(intToBinary(12))
console.log(intToBinary(546))
console.log(intToBinary(0))
console.log(intToBinary(125))
Works only with positive numbers.
An actual solution that logic can be implemented by any programming language:
If you sure it is positive only:
var a = 0;
var n = 12; // your input
var m = 1;
while(n) {
a = a + n%2*m;
n = Math.floor(n/2);
m = m*10;
}
console.log(n, ':', a) // 12 : 1100
If can negative or positive -
(n >>> 0).toString(2)
I’d like to see integers, positive or negative, in binary.
This is an old question and I think there are very nice solutions here but there is no explanation about the use of these clever solutions.
First, we need to understand that a number can be positive or negative.
Also, JavaScript provides a MAX_SAFE_INTEGER constant that has a value of 9007199254740991. The reasoning behind that number is that JavaScript uses double-precision floating-point format numbers as specified in IEEE 754 and can only safely represent integers between -(2^53 - 1) and 2^53 - 1.
So, now we know the range where numbers are "safe". Also, JavaScript ES6 has the built-in method Number.isSafeInteger() to check if a number is a safe integer.
Logically, if we want to represent a number n as binary, this number needs a length of 53 bits, but for better presentation lets use 7 groups of 8 bits = 56 bits and fill the left side with 0 or 1 based on its sign using the padStart function.
Next, we need to handle positive and negative numbers: positive numbers will add 0s to the left while negative numbers will add 1s. Also, negative numbers will need a two's-complement representation. We can easily fix this by adding Number.MAX_SAFE_INTEGER + 1 to the number.
For example, we want to represent -3 as binary, lets assume that Number.MAX_SAFE_INTEGER is 00000000 11111111 (255) then Number.MAX_SAFE_INTEGER + 1 will be 00000001 00000000 (256). Now lets add the number Number.MAX_SAFE_INTEGER + 1 - 3 this will be 00000000 11111101 (253) but as we said we will fill with the left side with 1 like this 11111111 11111101 (-3), this represent -3 in binary.
Another algorithm will be we add 1 to the number and invert the sign like this -(-3 + 1) = 2 this will be 00000000 00000010 (2). Now we invert every bit like this 11111111 11111101 (-3) again we have a binary representation of -3.
Here we have a working snippet of these algos:
function dec2binA(n) {
if (!Number.isSafeInteger(n)) throw new TypeError('n value must be a safe integer')
if (n > 2**31) throw 'number too large. number should not be greater than 2**31'
if (n < -1*(2**31)) throw 'number too far negative, number should not be lesser than 2**31'
const bin = n < 0 ? Number.MAX_SAFE_INTEGER + 1 + n : n
const signBit = n < 0 ? '1' : '0'
return parseInt(bin, 10).toString(2)
.padStart(56, signBit)
.replace(/\B(?=(.{8})+(?!.))/g, ' ')
}
function dec2binB(n) {
if (!Number.isSafeInteger(n)) throw new TypeError('n value must be a safe integer')
if (n > 2**31) throw 'number too large. number should not be greater than 2**31'
if (n < -1*(2**31)) throw 'number too far negative, number should not be lesser than 2**31'
const bin = n < 0 ? -(1 + n) : n
const signBit = n < 0 ? '1' : '0'
return parseInt(bin, 10).toString(2)
.replace(/[01]/g, d => +!+d)
.padStart(56, signBit)
.replace(/\B(?=(.{8})+(?!.))/g, ' ')
}
const a = -805306368
console.log(a)
console.log('dec2binA:', dec2binA(a))
console.log('dec2binB:', dec2binB(a))
const b = -3
console.log(b)
console.log('dec2binA:', dec2binA(b))
console.log('dec2binB:', dec2binB(b))
One more alternative
const decToBin = dec => {
let bin = '';
let f = false;
while (!f) {
bin = bin + (dec % 2);
dec = Math.trunc(dec / 2);
if (dec === 0 ) f = true;
}
return bin.split("").reverse().join("");
}
console.log(decToBin(0));
console.log(decToBin(1));
console.log(decToBin(2));
console.log(decToBin(3));
console.log(decToBin(4));
console.log(decToBin(5));
console.log(decToBin(6));
I used a different approach to come up with something that does this. I've decided to not use this code in my project, but I thought I'd leave it somewhere relevant in case it is useful for someone.
Doesn't use bit-shifting or two's complement coercion.
You choose the number of bits that comes out (it checks for valid values of '8', '16', '32', but I suppose you could change that)
You choose whether to treat it as a signed or unsigned integer.
It will check for range issues given the combination of signed/unsigned and number of bits, though you'll want to improve the error handling.
It also has the "reverse" version of the function which converts the bits back to the int. You'll need that since there's probably nothing else that will interpret this output :D
function intToBitString(input, size, unsigned) {
if ([8, 16, 32].indexOf(size) == -1) {
throw "invalid params";
}
var min = unsigned ? 0 : - (2 ** size / 2);
var limit = unsigned ? 2 ** size : 2 ** size / 2;
if (!Number.isInteger(input) || input < min || input >= limit) {
throw "out of range or not an int";
}
if (!unsigned) {
input += limit;
}
var binary = input.toString(2).replace(/^-/, '');
return binary.padStart(size, '0');
}
function bitStringToInt(input, size, unsigned) {
if ([8, 16, 32].indexOf(size) == -1) {
throw "invalid params";
}
input = parseInt(input, 2);
if (!unsigned) {
input -= 2 ** size / 2;
}
return input;
}
// EXAMPLES
var res;
console.log("(uint8)10");
res = intToBitString(10, 8, true);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(uint8)127");
res = intToBitString(127, 8, true);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(int8)127");
res = intToBitString(127, 8, false);
console.log("intToBitString(res, 8, false)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, false));
console.log("---");
console.log("(int8)-128");
res = intToBitString(-128, 8, false);
console.log("intToBitString(res, 8, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 8, true));
console.log("---");
console.log("(uint16)5000");
res = intToBitString(5000, 16, true);
console.log("intToBitString(res, 16, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 16, true));
console.log("---");
console.log("(uint32)5000");
res = intToBitString(5000, 32, true);
console.log("intToBitString(res, 32, true)");
console.log(res);
console.log("reverse:", bitStringToInt(res, 32, true));
console.log("---");
This is a method that I use. It's a very fast and concise method that works for whole numbers.
If you want, this method also works with BigInts. You just have to change each 1 to 1n.
// Assuming {num} is a whole number
function toBin(num){
let str = "";
do {
str = `${num & 1}${str}`;
num >>= 1;
} while(num);
return str
}
Explanation
This method, in a way, goes through all the bits of the number as if it's already a binary number.
It starts with an empty string, and then it prepends the last bit. num & 1 will return the last bit of the number (1 or 0). num >>= 1 then removes the last bit and makes the second-to-last bit the new last bit. The process is repeated until all the bits have been read.
Of course, this is an extreme simplification of what's actually going on. But this is how I generalize it.
This is my code:
var x = prompt("enter number", "7");
var i = 0;
var binaryvar = " ";
function add(n) {
if (n == 0) {
binaryvar = "0" + binaryvar;
}
else {
binaryvar = "1" + binaryvar;
}
}
function binary() {
while (i < 1) {
if (x == 1) {
add(1);
document.write(binaryvar);
break;
}
else {
if (x % 2 == 0) {
x = x / 2;
add(0);
}
else {
x = (x - 1) / 2;
add(1);
}
}
}
}
binary();
This is the solution . Its quite simple as a matter of fact
function binaries(num1){
var str = num1.toString(2)
return(console.log('The binary form of ' + num1 + ' is: ' + str))
}
binaries(3
)
/*
According to MDN, Number.prototype.toString() overrides
Object.prototype.toString() with the useful distinction that you can
pass in a single integer argument. This argument is an optional radix,
numbers 2 to 36 allowed.So in the example above, we’re passing in 2 to
get a string representation of the binary for the base 10 number 100,
i.e. 1100100.
*/
Related
How do I convert a string to an integer in JavaScript?
The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt()
var x = parseInt("1000", 10); // You want to use radix 10
// So you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
Unary plus
If your string is already in the form of an integer:
var x = +"1000";
floor()
If your string is or might be a float and you want an integer:
var x = Math.floor("1000.01"); // floor() automatically converts string to number
Or, if you're going to be using Math.floor several times:
var floor = Math.floor;
var x = floor("1000.01");
parseFloat()
If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
round()
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
var round = Math.round;
var x = round("1000"); // Equivalent to round("1000", 0)
Try parseInt function:
var number = parseInt("10");
But there is a problem. If you try to convert "010" using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
parseInt(string, radix)
Example:
var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
Note that parseInt ignores bad data after parsing anything valid.
This guid will parse as 51:
var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true
There are two main ways to convert a string to a number in JavaScript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g., unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
var parsed = parseInt("97", 10);
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn't recognise, which can be useful for parsing strings like "92px", but it's also somewhat dangerous, since it won't give you any kind of error on bad input, instead you'll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here's an example of it doing something different to what you want, and giving no indication that anything went wrong:
var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
It's good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn't specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g., 0xff. The standard actually changed with ECMAScript 5, so modern browsers no longer trigger octal when there's a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don't use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
var cast = Number("97");
This has different behavior to the parse methods (although it still ignores whitespace). It's more strict: if it doesn't understand the whole of the string than it returns NaN, so you can't use it for strings like 97px. Since you want a primitive number rather than a Number wrapper object, make sure you don't put new in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
Any bitwise operator (here I've done a bitwise or, but you could also do double negation as in an earlier answer or a bit shift) will convert the value to a 32 bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32 bits, it will wrap.
~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
To work correctly with larger numbers, you should use the rounding methods
Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
Bear in mind that coercion understands exponential notation and Infinity, so 2e2 is 200 rather than NaN, while the parse methods don't.
Custom
It's unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don't need support for Infinity, exponentials or leading whitespace. Depending on your use case, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use isNaN to make sure the number is not NaN (usually the only way you find out that the parse failed).
ParseInt() and + are different
parseInt("10.3456") // returns 10
+"10.3456" // returns 10.3456
Fastest
var x = "1000"*1;
Test
Here is little comparison of speed (macOS only)... :)
For Chrome, 'plus' and 'mul' are fastest (>700,000,00 op/sec), 'Math.floor' is slowest. For Firefox, 'plus' is slowest (!) 'mul' is fastest (>900,000,000 op/sec). In Safari 'parseInt' is fastest, 'number' is slowest (but results are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is 'mul' :)
You can run it on your browser by this link
https://jsperf.com/js-cast-str-to-number/1
I also tested var x = ~~"1000";. On Chrome and Safari, it is a little bit slower than var x = "1000"*1 (<1%), and on Firefox it is a little bit faster (<1%).
I use this way of converting string to number:
var str = "25"; // String
var number = str*1; // Number
So, when multiplying by 1, the value does not change, but JavaScript automatically returns a number.
But as it is shown below, this should be used if you are sure that the str is a number (or can be represented as a number), otherwise it will return NaN - not a number.
You can create simple function to use, e.g.,
function toNumber(str) {
return str*1;
}
Try parseInt.
var number = parseInt("10", 10); //number will have value of 10.
I love this trick:
~~"2.123"; //2
~~"5"; //5
The double bitwise negative drops off anything after the decimal point AND converts it to a number format. I've been told it's slightly faster than calling functions and whatnot, but I'm not entirely convinced.
Another method I just saw here (a question about the JavaScript >>> operator, which is a zero-fill right shift) which shows that shifting a number by 0 with this operator converts the number to a uint32 which is nice if you also want it unsigned. Again, this converts to an unsigned integer, which can lead to strange behaviors if you use a signed number.
"-2.123" >>> 0; // 4294967294
"2.123" >>> 0; // 2
"-5" >>> 0; // 4294967291
"5" >>> 0; // 5
In JavaScript, you can do the following:
ParseInt
parseInt("10.5") // Returns 10
Multiplying with 1
var s = "10";
s = s*1; // Returns 10
Using the unary operator (+)
var s = "10";
s = +s; // Returns 10
Using a bitwise operator
(Note: It starts to break after 2140000000. Example: ~~"2150000000" = -2144967296)
var s = "10.5";
s = ~~s; // Returns 10
Using Math.floor() or Math.ceil()
var s = "10";
s = Math.floor(s) || Math.ceil(s); // Returns 10
Please see the below example. It will help answer your question.
Example Result
parseInt("4") 4
parseInt("5aaa") 5
parseInt("4.33333") 4
parseInt("aaa"); NaN (means "Not a Number")
By using parseint function, it will only give op of integer present and not the string.
Beware if you use parseInt to convert a float in scientific notation!
For example:
parseInt("5.6e-14")
will result in
5
instead of
0
Also as a side note: MooTools has the function toInt() which is used on any native string (or float (or integer)).
"2".toInt() // 2
"2px".toInt() // 2
2.toInt() // 2
We can use +(stringOfNumber) instead of using parseInt(stringOfNumber).
Example: +("21") returns int of 21, like the parseInt("21").
We can use this unary "+" operator for parsing float too...
To convert a String into Integer, I recommend using parseFloat and not parseInt. Here's why:
Using parseFloat:
parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
Using parseInt:
parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value.
There are many ways in JavaScript to convert a string to a number value... All are simple and handy. Choose the way which one works for you:
var num = Number("999.5"); //999.5
var num = parseInt("999.5", 10); //999
var num = parseFloat("999.5"); //999.5
var num = +"999.5"; //999.5
Also, any Math operation converts them to number, for example...
var num = "999.5" / 1; //999.5
var num = "999.5" * 1; //999.5
var num = "999.5" - 1 + 1; //999.5
var num = "999.5" - 0; //999.5
var num = Math.floor("999.5"); //999
var num = ~~"999.5"; //999
My prefer way is using + sign, which is the elegant way to convert a string to number in JavaScript.
Try str - 0 to convert string to number.
> str = '0'
> str - 0
0
> str = '123'
> str - 0
123
> str = '-12'
> str - 0
-12
> str = 'asdf'
> str - 0
NaN
> str = '12.34'
> str - 0
12.34
Here are two links to compare the performance of several ways to convert string to int
https://jsperf.com/number-vs-parseint-vs-plus
http://phrogz.net/js/string_to_number.html
Here is the easiest solution
let myNumber = "123" | 0;
More easy solution
let myNumber = +"123";
In my opinion, no answer covers all edge cases as parsing a float should result in an error.
function parseInteger(value) {
if(value === '') return NaN;
const number = Number(value);
return Number.isInteger(number) ? number : NaN;
}
parseInteger("4") // 4
parseInteger("5aaa") // NaN
parseInteger("4.33333") // NaN
parseInteger("aaa"); // NaN
The easiest way would be to use + like this
const strTen = "10"
const numTen = +strTen // string to number conversion
console.log(typeof strTen) // string
console.log(typeof numTen) // number
I actually needed to "save" a string as an integer, for a binding between C and JavaScript, so I convert the string into an integer value:
/*
Examples:
int2str( str2int("test") ) == "test" // true
int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
Limitations:
maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
var ret = 0;
var len = the_str.length;
if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) << 0;
if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) << 8;
if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
return ret;
}
function int2str(the_int) {
var tmp = [
(the_int & 0x000000ff) >> 0,
(the_int & 0x0000ff00) >> 8,
(the_int & 0x00ff0000) >> 16,
(the_int & 0xff000000) >> 24
];
var ret = "";
for (var i=0; i<4; i++) {
if (tmp[i] == 0)
break;
ret += String.fromCharCode(tmp[i]);
}
return ret;
}
String to Number in JavaScript:
Unary + (most recommended)
+numStr is easy to use and has better performance compared with others
Supports both integers and decimals
console.log(+'123.45') // => 123.45
Some other options:
Parsing Strings:
parseInt(numStr) for integers
parseFloat(numStr) for both integers and decimals
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
JavaScript Functions
Math functions like round(numStr), floor(numStr), ceil(numStr) for integers
Number(numStr) for both integers and decimals
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
Unary Operators
All basic unary operators, +numStr, numStr-0, 1*numStr, numStr*1, and numStr/1
All support both integers and decimals
Be cautious about numStr+0. It returns a string.
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // =>3.3
console.log('123.123'+0, typeof ('123.123' + 0)) // => 123.1230 string
Bitwise Operators
Two tilde ~~numStr or left shift 0, numStr<<0
Supports only integers, but not decimals
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
// Parsing
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
// Function
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
// Unary
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // => 3.3
console.log('123.123'+0, typeof ('123.123'+0)) // => 123.1230 string
// Bitwise
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
function parseIntSmarter(str) {
// ParseInt is bad because it returns 22 for "22thisendsintext"
// Number() is returns NaN if it ends in non-numbers, but it returns 0 for empty or whitespace strings.
return isNaN(Number(str)) ? NaN : parseInt(str, 10);
}
You can use plus.
For example:
var personAge = '24';
var personAge1 = (+personAge)
then you can see the new variable's type bytypeof personAge1 ; which is number.
Summing the multiplication of digits with their respective power of ten:
i.e: 123 = 100+20+3 = 1100 + 2+10 + 31 = 1*(10^2) + 2*(10^1) + 3*(10^0)
function atoi(array) {
// Use exp as (length - i), other option would be
// to reverse the array.
// Multiply a[i] * 10^(exp) and sum
let sum = 0;
for (let i = 0; i < array.length; i++) {
let exp = array.length - (i+1);
let value = array[i] * Math.pow(10, exp);
sum += value;
}
return sum;
}
The safest way to ensure you get a valid integer:
let integer = (parseInt(value, 10) || 0);
Examples:
// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0
Another option is to double XOR the value with itself:
var i = 12.34;
console.log('i = ' + i);
console.log('i ⊕ i ⊕ i = ' + (i ^ i ^ i));
This will output:
i = 12.34
i ⊕ i ⊕ i = 12
I only added one plus(+) before string and that was solution!
+"052254" // 52254
Number()
Number(" 200.12 ") // Returns 200.12
Number("200.12") // Returns 200.12
Number("200") // Returns 200
parseInt()
parseInt(" 200.12 ") // Return 200
parseInt("200.12") // Return 200
parseInt("200") // Return 200
parseInt("Text information") // Returns NaN
parseFloat()
It will return the first number
parseFloat("200 400") // Returns 200
parseFloat("200") // Returns 200
parseFloat("Text information") // Returns NaN
parseFloat("200.10") // Return 200.10
Math.floor()
Round a number to the nearest integer
Math.floor(" 200.12 ") // Return 200
Math.floor("200.12") // Return 200
Math.floor("200") // Return 200
function doSth(){
var a = document.getElementById('input').value;
document.getElementById('number').innerHTML = toNumber(a) + 1;
}
function toNumber(str){
return +str;
}
<input id="input" type="text">
<input onclick="doSth()" type="submit">
<span id="number"></span>
This (probably) isn't the best solution for parsing an integer, but if you need to "extract" one, for example:
"1a2b3c" === 123
"198some text2hello world!30" === 198230
// ...
this would work (only for integers):
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (!isNaN(str[i - 1])) {
result += parseInt(str[i - 1]) * factor
factor *= 10
}
}
return result
}
console.log(extractInteger(str))
Of course, this would also work for parsing an integer, but would be slower than other methods.
You could also parse integers with this method and return NaN if the string isn't a number, but I don't see why you'd want to since this relies on parseInt internally and parseInt is probably faster.
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (isNaN(str[i - 1])) return NaN
result += parseInt(str[i - 1]) * factor
factor *= 10
}
return result
}
console.log(extractInteger(str))
How do I convert a string to an integer in JavaScript?
The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt()
var x = parseInt("1000", 10); // You want to use radix 10
// So you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
Unary plus
If your string is already in the form of an integer:
var x = +"1000";
floor()
If your string is or might be a float and you want an integer:
var x = Math.floor("1000.01"); // floor() automatically converts string to number
Or, if you're going to be using Math.floor several times:
var floor = Math.floor;
var x = floor("1000.01");
parseFloat()
If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
round()
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
var round = Math.round;
var x = round("1000"); // Equivalent to round("1000", 0)
Try parseInt function:
var number = parseInt("10");
But there is a problem. If you try to convert "010" using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
parseInt(string, radix)
Example:
var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
Note that parseInt ignores bad data after parsing anything valid.
This guid will parse as 51:
var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true
There are two main ways to convert a string to a number in JavaScript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g., unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
var parsed = parseInt("97", 10);
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn't recognise, which can be useful for parsing strings like "92px", but it's also somewhat dangerous, since it won't give you any kind of error on bad input, instead you'll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here's an example of it doing something different to what you want, and giving no indication that anything went wrong:
var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
It's good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn't specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g., 0xff. The standard actually changed with ECMAScript 5, so modern browsers no longer trigger octal when there's a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don't use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
var cast = Number("97");
This has different behavior to the parse methods (although it still ignores whitespace). It's more strict: if it doesn't understand the whole of the string than it returns NaN, so you can't use it for strings like 97px. Since you want a primitive number rather than a Number wrapper object, make sure you don't put new in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
Any bitwise operator (here I've done a bitwise or, but you could also do double negation as in an earlier answer or a bit shift) will convert the value to a 32 bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32 bits, it will wrap.
~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
To work correctly with larger numbers, you should use the rounding methods
Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
Bear in mind that coercion understands exponential notation and Infinity, so 2e2 is 200 rather than NaN, while the parse methods don't.
Custom
It's unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don't need support for Infinity, exponentials or leading whitespace. Depending on your use case, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use isNaN to make sure the number is not NaN (usually the only way you find out that the parse failed).
ParseInt() and + are different
parseInt("10.3456") // returns 10
+"10.3456" // returns 10.3456
Fastest
var x = "1000"*1;
Test
Here is little comparison of speed (macOS only)... :)
For Chrome, 'plus' and 'mul' are fastest (>700,000,00 op/sec), 'Math.floor' is slowest. For Firefox, 'plus' is slowest (!) 'mul' is fastest (>900,000,000 op/sec). In Safari 'parseInt' is fastest, 'number' is slowest (but results are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is 'mul' :)
You can run it on your browser by this link
https://jsperf.com/js-cast-str-to-number/1
I also tested var x = ~~"1000";. On Chrome and Safari, it is a little bit slower than var x = "1000"*1 (<1%), and on Firefox it is a little bit faster (<1%).
I use this way of converting string to number:
var str = "25"; // String
var number = str*1; // Number
So, when multiplying by 1, the value does not change, but JavaScript automatically returns a number.
But as it is shown below, this should be used if you are sure that the str is a number (or can be represented as a number), otherwise it will return NaN - not a number.
You can create simple function to use, e.g.,
function toNumber(str) {
return str*1;
}
Try parseInt.
var number = parseInt("10", 10); //number will have value of 10.
I love this trick:
~~"2.123"; //2
~~"5"; //5
The double bitwise negative drops off anything after the decimal point AND converts it to a number format. I've been told it's slightly faster than calling functions and whatnot, but I'm not entirely convinced.
Another method I just saw here (a question about the JavaScript >>> operator, which is a zero-fill right shift) which shows that shifting a number by 0 with this operator converts the number to a uint32 which is nice if you also want it unsigned. Again, this converts to an unsigned integer, which can lead to strange behaviors if you use a signed number.
"-2.123" >>> 0; // 4294967294
"2.123" >>> 0; // 2
"-5" >>> 0; // 4294967291
"5" >>> 0; // 5
In JavaScript, you can do the following:
ParseInt
parseInt("10.5") // Returns 10
Multiplying with 1
var s = "10";
s = s*1; // Returns 10
Using the unary operator (+)
var s = "10";
s = +s; // Returns 10
Using a bitwise operator
(Note: It starts to break after 2140000000. Example: ~~"2150000000" = -2144967296)
var s = "10.5";
s = ~~s; // Returns 10
Using Math.floor() or Math.ceil()
var s = "10";
s = Math.floor(s) || Math.ceil(s); // Returns 10
Please see the below example. It will help answer your question.
Example Result
parseInt("4") 4
parseInt("5aaa") 5
parseInt("4.33333") 4
parseInt("aaa"); NaN (means "Not a Number")
By using parseint function, it will only give op of integer present and not the string.
Beware if you use parseInt to convert a float in scientific notation!
For example:
parseInt("5.6e-14")
will result in
5
instead of
0
Also as a side note: MooTools has the function toInt() which is used on any native string (or float (or integer)).
"2".toInt() // 2
"2px".toInt() // 2
2.toInt() // 2
We can use +(stringOfNumber) instead of using parseInt(stringOfNumber).
Example: +("21") returns int of 21, like the parseInt("21").
We can use this unary "+" operator for parsing float too...
To convert a String into Integer, I recommend using parseFloat and not parseInt. Here's why:
Using parseFloat:
parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
Using parseInt:
parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value.
There are many ways in JavaScript to convert a string to a number value... All are simple and handy. Choose the way which one works for you:
var num = Number("999.5"); //999.5
var num = parseInt("999.5", 10); //999
var num = parseFloat("999.5"); //999.5
var num = +"999.5"; //999.5
Also, any Math operation converts them to number, for example...
var num = "999.5" / 1; //999.5
var num = "999.5" * 1; //999.5
var num = "999.5" - 1 + 1; //999.5
var num = "999.5" - 0; //999.5
var num = Math.floor("999.5"); //999
var num = ~~"999.5"; //999
My prefer way is using + sign, which is the elegant way to convert a string to number in JavaScript.
Try str - 0 to convert string to number.
> str = '0'
> str - 0
0
> str = '123'
> str - 0
123
> str = '-12'
> str - 0
-12
> str = 'asdf'
> str - 0
NaN
> str = '12.34'
> str - 0
12.34
Here are two links to compare the performance of several ways to convert string to int
https://jsperf.com/number-vs-parseint-vs-plus
http://phrogz.net/js/string_to_number.html
Here is the easiest solution
let myNumber = "123" | 0;
More easy solution
let myNumber = +"123";
In my opinion, no answer covers all edge cases as parsing a float should result in an error.
function parseInteger(value) {
if(value === '') return NaN;
const number = Number(value);
return Number.isInteger(number) ? number : NaN;
}
parseInteger("4") // 4
parseInteger("5aaa") // NaN
parseInteger("4.33333") // NaN
parseInteger("aaa"); // NaN
The easiest way would be to use + like this
const strTen = "10"
const numTen = +strTen // string to number conversion
console.log(typeof strTen) // string
console.log(typeof numTen) // number
I actually needed to "save" a string as an integer, for a binding between C and JavaScript, so I convert the string into an integer value:
/*
Examples:
int2str( str2int("test") ) == "test" // true
int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
Limitations:
maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
var ret = 0;
var len = the_str.length;
if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) << 0;
if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) << 8;
if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
return ret;
}
function int2str(the_int) {
var tmp = [
(the_int & 0x000000ff) >> 0,
(the_int & 0x0000ff00) >> 8,
(the_int & 0x00ff0000) >> 16,
(the_int & 0xff000000) >> 24
];
var ret = "";
for (var i=0; i<4; i++) {
if (tmp[i] == 0)
break;
ret += String.fromCharCode(tmp[i]);
}
return ret;
}
String to Number in JavaScript:
Unary + (most recommended)
+numStr is easy to use and has better performance compared with others
Supports both integers and decimals
console.log(+'123.45') // => 123.45
Some other options:
Parsing Strings:
parseInt(numStr) for integers
parseFloat(numStr) for both integers and decimals
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
JavaScript Functions
Math functions like round(numStr), floor(numStr), ceil(numStr) for integers
Number(numStr) for both integers and decimals
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
Unary Operators
All basic unary operators, +numStr, numStr-0, 1*numStr, numStr*1, and numStr/1
All support both integers and decimals
Be cautious about numStr+0. It returns a string.
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // =>3.3
console.log('123.123'+0, typeof ('123.123' + 0)) // => 123.1230 string
Bitwise Operators
Two tilde ~~numStr or left shift 0, numStr<<0
Supports only integers, but not decimals
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
// Parsing
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
// Function
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
// Unary
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // => 3.3
console.log('123.123'+0, typeof ('123.123'+0)) // => 123.1230 string
// Bitwise
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
function parseIntSmarter(str) {
// ParseInt is bad because it returns 22 for "22thisendsintext"
// Number() is returns NaN if it ends in non-numbers, but it returns 0 for empty or whitespace strings.
return isNaN(Number(str)) ? NaN : parseInt(str, 10);
}
You can use plus.
For example:
var personAge = '24';
var personAge1 = (+personAge)
then you can see the new variable's type bytypeof personAge1 ; which is number.
Summing the multiplication of digits with their respective power of ten:
i.e: 123 = 100+20+3 = 1100 + 2+10 + 31 = 1*(10^2) + 2*(10^1) + 3*(10^0)
function atoi(array) {
// Use exp as (length - i), other option would be
// to reverse the array.
// Multiply a[i] * 10^(exp) and sum
let sum = 0;
for (let i = 0; i < array.length; i++) {
let exp = array.length - (i+1);
let value = array[i] * Math.pow(10, exp);
sum += value;
}
return sum;
}
The safest way to ensure you get a valid integer:
let integer = (parseInt(value, 10) || 0);
Examples:
// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0
Another option is to double XOR the value with itself:
var i = 12.34;
console.log('i = ' + i);
console.log('i ⊕ i ⊕ i = ' + (i ^ i ^ i));
This will output:
i = 12.34
i ⊕ i ⊕ i = 12
I only added one plus(+) before string and that was solution!
+"052254" // 52254
Number()
Number(" 200.12 ") // Returns 200.12
Number("200.12") // Returns 200.12
Number("200") // Returns 200
parseInt()
parseInt(" 200.12 ") // Return 200
parseInt("200.12") // Return 200
parseInt("200") // Return 200
parseInt("Text information") // Returns NaN
parseFloat()
It will return the first number
parseFloat("200 400") // Returns 200
parseFloat("200") // Returns 200
parseFloat("Text information") // Returns NaN
parseFloat("200.10") // Return 200.10
Math.floor()
Round a number to the nearest integer
Math.floor(" 200.12 ") // Return 200
Math.floor("200.12") // Return 200
Math.floor("200") // Return 200
function doSth(){
var a = document.getElementById('input').value;
document.getElementById('number').innerHTML = toNumber(a) + 1;
}
function toNumber(str){
return +str;
}
<input id="input" type="text">
<input onclick="doSth()" type="submit">
<span id="number"></span>
This (probably) isn't the best solution for parsing an integer, but if you need to "extract" one, for example:
"1a2b3c" === 123
"198some text2hello world!30" === 198230
// ...
this would work (only for integers):
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (!isNaN(str[i - 1])) {
result += parseInt(str[i - 1]) * factor
factor *= 10
}
}
return result
}
console.log(extractInteger(str))
Of course, this would also work for parsing an integer, but would be slower than other methods.
You could also parse integers with this method and return NaN if the string isn't a number, but I don't see why you'd want to since this relies on parseInt internally and parseInt is probably faster.
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (isNaN(str[i - 1])) return NaN
result += parseInt(str[i - 1]) * factor
factor *= 10
}
return result
}
console.log(extractInteger(str))
How do I convert a string to an integer in JavaScript?
The simplest way would be to use the native Number function:
var x = Number("1000")
If that doesn't work for you, then there are the parseInt, unary plus, parseFloat with floor, and Math.round methods.
parseInt()
var x = parseInt("1000", 10); // You want to use radix 10
// So you get a decimal number even with a leading 0 and an old browser ([IE8, Firefox 20, Chrome 22 and older][1])
Unary plus
If your string is already in the form of an integer:
var x = +"1000";
floor()
If your string is or might be a float and you want an integer:
var x = Math.floor("1000.01"); // floor() automatically converts string to number
Or, if you're going to be using Math.floor several times:
var floor = Math.floor;
var x = floor("1000.01");
parseFloat()
If you're the type who forgets to put the radix in when you call parseInt, you can use parseFloat and round it however you like. Here I use floor.
var floor = Math.floor;
var x = floor(parseFloat("1000.01"));
round()
Interestingly, Math.round (like Math.floor) will do a string to number conversion, so if you want the number rounded (or if you have an integer in the string), this is a great way, maybe my favorite:
var round = Math.round;
var x = round("1000"); // Equivalent to round("1000", 0)
Try parseInt function:
var number = parseInt("10");
But there is a problem. If you try to convert "010" using parseInt function, it detects as octal number, and will return number 8. So, you need to specify a radix (from 2 to 36). In this case base 10.
parseInt(string, radix)
Example:
var result = parseInt("010", 10) == 10; // Returns true
var result = parseInt("010") == 10; // Returns false
Note that parseInt ignores bad data after parsing anything valid.
This guid will parse as 51:
var result = parseInt('51e3daf6-b521-446a-9f5b-a1bb4d8bac36', 10) == 51; // Returns true
There are two main ways to convert a string to a number in JavaScript. One way is to parse it and the other way is to change its type to a Number. All of the tricks in the other answers (e.g., unary plus) involve implicitly coercing the type of the string to a number. You can also do the same thing explicitly with the Number function.
Parsing
var parsed = parseInt("97", 10);
parseInt and parseFloat are the two functions used for parsing strings to numbers. Parsing will stop silently if it hits a character it doesn't recognise, which can be useful for parsing strings like "92px", but it's also somewhat dangerous, since it won't give you any kind of error on bad input, instead you'll get back NaN unless the string starts with a number. Whitespace at the beginning of the string is ignored. Here's an example of it doing something different to what you want, and giving no indication that anything went wrong:
var widgetsSold = parseInt("97,800", 10); // widgetsSold is now 97
It's good practice to always specify the radix as the second argument. In older browsers, if the string started with a 0, it would be interpreted as octal if the radix wasn't specified which took a lot of people by surprise. The behaviour for hexadecimal is triggered by having the string start with 0x if no radix is specified, e.g., 0xff. The standard actually changed with ECMAScript 5, so modern browsers no longer trigger octal when there's a leading 0 if no radix has been specified. parseInt understands radixes up to base 36, in which case both upper and lower case letters are treated as equivalent.
Changing the Type of a String to a Number
All of the other tricks mentioned above that don't use parseInt, involve implicitly coercing the string into a number. I prefer to do this explicitly,
var cast = Number("97");
This has different behavior to the parse methods (although it still ignores whitespace). It's more strict: if it doesn't understand the whole of the string than it returns NaN, so you can't use it for strings like 97px. Since you want a primitive number rather than a Number wrapper object, make sure you don't put new in front of the Number function.
Obviously, converting to a Number gives you a value that might be a float rather than an integer, so if you want an integer, you need to modify it. There are a few ways of doing this:
var rounded = Math.floor(Number("97.654")); // other options are Math.ceil, Math.round
var fixed = Number("97.654").toFixed(0); // rounded rather than truncated
var bitwised = Number("97.654")|0; // do not use for large numbers
Any bitwise operator (here I've done a bitwise or, but you could also do double negation as in an earlier answer or a bit shift) will convert the value to a 32 bit integer, and most of them will convert to a signed integer. Note that this will not do want you want for large integers. If the integer cannot be represented in 32 bits, it will wrap.
~~"3000000000.654" === -1294967296
// This is the same as
Number("3000000000.654")|0
"3000000000.654" >>> 0 === 3000000000 // unsigned right shift gives you an extra bit
"300000000000.654" >>> 0 === 3647256576 // but still fails with larger numbers
To work correctly with larger numbers, you should use the rounding methods
Math.floor("3000000000.654") === 3000000000
// This is the same as
Math.floor(Number("3000000000.654"))
Bear in mind that coercion understands exponential notation and Infinity, so 2e2 is 200 rather than NaN, while the parse methods don't.
Custom
It's unlikely that either of these methods do exactly what you want. For example, usually I would want an error thrown if parsing fails, and I don't need support for Infinity, exponentials or leading whitespace. Depending on your use case, sometimes it makes sense to write a custom conversion function.
Always check that the output of Number or one of the parse methods is the sort of number you expect. You will almost certainly want to use isNaN to make sure the number is not NaN (usually the only way you find out that the parse failed).
ParseInt() and + are different
parseInt("10.3456") // returns 10
+"10.3456" // returns 10.3456
Fastest
var x = "1000"*1;
Test
Here is little comparison of speed (macOS only)... :)
For Chrome, 'plus' and 'mul' are fastest (>700,000,00 op/sec), 'Math.floor' is slowest. For Firefox, 'plus' is slowest (!) 'mul' is fastest (>900,000,000 op/sec). In Safari 'parseInt' is fastest, 'number' is slowest (but results are quite similar, >13,000,000 <31,000,000). So Safari for cast string to int is more than 10x slower than other browsers. So the winner is 'mul' :)
You can run it on your browser by this link
https://jsperf.com/js-cast-str-to-number/1
I also tested var x = ~~"1000";. On Chrome and Safari, it is a little bit slower than var x = "1000"*1 (<1%), and on Firefox it is a little bit faster (<1%).
I use this way of converting string to number:
var str = "25"; // String
var number = str*1; // Number
So, when multiplying by 1, the value does not change, but JavaScript automatically returns a number.
But as it is shown below, this should be used if you are sure that the str is a number (or can be represented as a number), otherwise it will return NaN - not a number.
You can create simple function to use, e.g.,
function toNumber(str) {
return str*1;
}
Try parseInt.
var number = parseInt("10", 10); //number will have value of 10.
I love this trick:
~~"2.123"; //2
~~"5"; //5
The double bitwise negative drops off anything after the decimal point AND converts it to a number format. I've been told it's slightly faster than calling functions and whatnot, but I'm not entirely convinced.
Another method I just saw here (a question about the JavaScript >>> operator, which is a zero-fill right shift) which shows that shifting a number by 0 with this operator converts the number to a uint32 which is nice if you also want it unsigned. Again, this converts to an unsigned integer, which can lead to strange behaviors if you use a signed number.
"-2.123" >>> 0; // 4294967294
"2.123" >>> 0; // 2
"-5" >>> 0; // 4294967291
"5" >>> 0; // 5
In JavaScript, you can do the following:
ParseInt
parseInt("10.5") // Returns 10
Multiplying with 1
var s = "10";
s = s*1; // Returns 10
Using the unary operator (+)
var s = "10";
s = +s; // Returns 10
Using a bitwise operator
(Note: It starts to break after 2140000000. Example: ~~"2150000000" = -2144967296)
var s = "10.5";
s = ~~s; // Returns 10
Using Math.floor() or Math.ceil()
var s = "10";
s = Math.floor(s) || Math.ceil(s); // Returns 10
Please see the below example. It will help answer your question.
Example Result
parseInt("4") 4
parseInt("5aaa") 5
parseInt("4.33333") 4
parseInt("aaa"); NaN (means "Not a Number")
By using parseint function, it will only give op of integer present and not the string.
Beware if you use parseInt to convert a float in scientific notation!
For example:
parseInt("5.6e-14")
will result in
5
instead of
0
Also as a side note: MooTools has the function toInt() which is used on any native string (or float (or integer)).
"2".toInt() // 2
"2px".toInt() // 2
2.toInt() // 2
We can use +(stringOfNumber) instead of using parseInt(stringOfNumber).
Example: +("21") returns int of 21, like the parseInt("21").
We can use this unary "+" operator for parsing float too...
To convert a String into Integer, I recommend using parseFloat and not parseInt. Here's why:
Using parseFloat:
parseFloat('2.34cms') //Output: 2.34
parseFloat('12.5') //Output: 12.5
parseFloat('012.3') //Output: 12.3
Using parseInt:
parseInt('2.34cms') //Output: 2
parseInt('12.5') //Output: 12
parseInt('012.3') //Output: 12
So if you have noticed parseInt discards the values after the decimals, whereas parseFloat lets you work with floating point numbers and hence more suitable if you want to retain the values after decimals. Use parseInt if and only if you are sure that you want the integer value.
There are many ways in JavaScript to convert a string to a number value... All are simple and handy. Choose the way which one works for you:
var num = Number("999.5"); //999.5
var num = parseInt("999.5", 10); //999
var num = parseFloat("999.5"); //999.5
var num = +"999.5"; //999.5
Also, any Math operation converts them to number, for example...
var num = "999.5" / 1; //999.5
var num = "999.5" * 1; //999.5
var num = "999.5" - 1 + 1; //999.5
var num = "999.5" - 0; //999.5
var num = Math.floor("999.5"); //999
var num = ~~"999.5"; //999
My prefer way is using + sign, which is the elegant way to convert a string to number in JavaScript.
Try str - 0 to convert string to number.
> str = '0'
> str - 0
0
> str = '123'
> str - 0
123
> str = '-12'
> str - 0
-12
> str = 'asdf'
> str - 0
NaN
> str = '12.34'
> str - 0
12.34
Here are two links to compare the performance of several ways to convert string to int
https://jsperf.com/number-vs-parseint-vs-plus
http://phrogz.net/js/string_to_number.html
Here is the easiest solution
let myNumber = "123" | 0;
More easy solution
let myNumber = +"123";
In my opinion, no answer covers all edge cases as parsing a float should result in an error.
function parseInteger(value) {
if(value === '') return NaN;
const number = Number(value);
return Number.isInteger(number) ? number : NaN;
}
parseInteger("4") // 4
parseInteger("5aaa") // NaN
parseInteger("4.33333") // NaN
parseInteger("aaa"); // NaN
The easiest way would be to use + like this
const strTen = "10"
const numTen = +strTen // string to number conversion
console.log(typeof strTen) // string
console.log(typeof numTen) // number
I actually needed to "save" a string as an integer, for a binding between C and JavaScript, so I convert the string into an integer value:
/*
Examples:
int2str( str2int("test") ) == "test" // true
int2str( str2int("t€st") ) // "t¬st", because "€".charCodeAt(0) is 8364, will be AND'ed with 0xff
Limitations:
maximum 4 characters, so it fits into an integer
*/
function str2int(the_str) {
var ret = 0;
var len = the_str.length;
if (len >= 1) ret += (the_str.charCodeAt(0) & 0xff) << 0;
if (len >= 2) ret += (the_str.charCodeAt(1) & 0xff) << 8;
if (len >= 3) ret += (the_str.charCodeAt(2) & 0xff) << 16;
if (len >= 4) ret += (the_str.charCodeAt(3) & 0xff) << 24;
return ret;
}
function int2str(the_int) {
var tmp = [
(the_int & 0x000000ff) >> 0,
(the_int & 0x0000ff00) >> 8,
(the_int & 0x00ff0000) >> 16,
(the_int & 0xff000000) >> 24
];
var ret = "";
for (var i=0; i<4; i++) {
if (tmp[i] == 0)
break;
ret += String.fromCharCode(tmp[i]);
}
return ret;
}
String to Number in JavaScript:
Unary + (most recommended)
+numStr is easy to use and has better performance compared with others
Supports both integers and decimals
console.log(+'123.45') // => 123.45
Some other options:
Parsing Strings:
parseInt(numStr) for integers
parseFloat(numStr) for both integers and decimals
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
JavaScript Functions
Math functions like round(numStr), floor(numStr), ceil(numStr) for integers
Number(numStr) for both integers and decimals
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
Unary Operators
All basic unary operators, +numStr, numStr-0, 1*numStr, numStr*1, and numStr/1
All support both integers and decimals
Be cautious about numStr+0. It returns a string.
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // =>3.3
console.log('123.123'+0, typeof ('123.123' + 0)) // => 123.1230 string
Bitwise Operators
Two tilde ~~numStr or left shift 0, numStr<<0
Supports only integers, but not decimals
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
// Parsing
console.log(parseInt('123.456')) // => 123
console.log(parseFloat('123')) // => 123
// Function
console.log(Math.floor('123')) // => 123
console.log(Math.round('123.456')) // => 123
console.log(Math.ceil('123.454')) // => 124
console.log(Number('123.123')) // => 123.123
// Unary
console.log(+'123') // => 123
console.log('002'-0) // => 2
console.log(1*'5') // => 5
console.log('7.7'*1) // => 7.7
console.log(3.3/1) // => 3.3
console.log('123.123'+0, typeof ('123.123'+0)) // => 123.1230 string
// Bitwise
console.log(~~'123') // => 123
console.log('0123'<<0) // => 123
console.log(~~'123.123') // => 123
console.log('123.123'<<0) // => 123
function parseIntSmarter(str) {
// ParseInt is bad because it returns 22 for "22thisendsintext"
// Number() is returns NaN if it ends in non-numbers, but it returns 0 for empty or whitespace strings.
return isNaN(Number(str)) ? NaN : parseInt(str, 10);
}
You can use plus.
For example:
var personAge = '24';
var personAge1 = (+personAge)
then you can see the new variable's type bytypeof personAge1 ; which is number.
Summing the multiplication of digits with their respective power of ten:
i.e: 123 = 100+20+3 = 1100 + 2+10 + 31 = 1*(10^2) + 2*(10^1) + 3*(10^0)
function atoi(array) {
// Use exp as (length - i), other option would be
// to reverse the array.
// Multiply a[i] * 10^(exp) and sum
let sum = 0;
for (let i = 0; i < array.length; i++) {
let exp = array.length - (i+1);
let value = array[i] * Math.pow(10, exp);
sum += value;
}
return sum;
}
The safest way to ensure you get a valid integer:
let integer = (parseInt(value, 10) || 0);
Examples:
// Example 1 - Invalid value:
let value = null;
let integer = (parseInt(value, 10) || 0);
// => integer = 0
// Example 2 - Valid value:
let value = "1230.42";
let integer = (parseInt(value, 10) || 0);
// => integer = 1230
// Example 3 - Invalid value:
let value = () => { return 412 };
let integer = (parseInt(value, 10) || 0);
// => integer = 0
Another option is to double XOR the value with itself:
var i = 12.34;
console.log('i = ' + i);
console.log('i ⊕ i ⊕ i = ' + (i ^ i ^ i));
This will output:
i = 12.34
i ⊕ i ⊕ i = 12
I only added one plus(+) before string and that was solution!
+"052254" // 52254
Number()
Number(" 200.12 ") // Returns 200.12
Number("200.12") // Returns 200.12
Number("200") // Returns 200
parseInt()
parseInt(" 200.12 ") // Return 200
parseInt("200.12") // Return 200
parseInt("200") // Return 200
parseInt("Text information") // Returns NaN
parseFloat()
It will return the first number
parseFloat("200 400") // Returns 200
parseFloat("200") // Returns 200
parseFloat("Text information") // Returns NaN
parseFloat("200.10") // Return 200.10
Math.floor()
Round a number to the nearest integer
Math.floor(" 200.12 ") // Return 200
Math.floor("200.12") // Return 200
Math.floor("200") // Return 200
function doSth(){
var a = document.getElementById('input').value;
document.getElementById('number').innerHTML = toNumber(a) + 1;
}
function toNumber(str){
return +str;
}
<input id="input" type="text">
<input onclick="doSth()" type="submit">
<span id="number"></span>
This (probably) isn't the best solution for parsing an integer, but if you need to "extract" one, for example:
"1a2b3c" === 123
"198some text2hello world!30" === 198230
// ...
this would work (only for integers):
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (!isNaN(str[i - 1])) {
result += parseInt(str[i - 1]) * factor
factor *= 10
}
}
return result
}
console.log(extractInteger(str))
Of course, this would also work for parsing an integer, but would be slower than other methods.
You could also parse integers with this method and return NaN if the string isn't a number, but I don't see why you'd want to since this relies on parseInt internally and parseInt is probably faster.
var str = '3a9b0c3d2e9f8g'
function extractInteger(str) {
var result = 0;
var factor = 1
for (var i = str.length; i > 0; i--) {
if (isNaN(str[i - 1])) return NaN
result += parseInt(str[i - 1]) * factor
factor *= 10
}
return result
}
console.log(extractInteger(str))
I have a 64 bit unsigned integer I need to represent in PostgreSQL. I've broken it down into two 32 bit unsigned integers, high and low. To allow Postgres to accept it, I need to convert high and low to a string representing a signed 64 bit integer.
How can I go about converting two 32 bit unsigned integers to a string representing in decimal a signed 64 bit integer?
I've done exactly this in Javascript in a quick'n'dirty-but-works'n'fast manner at: Int64HighLowToFromString, using 53-bit mantissa double precision arithmetic and 32-bit bit operations, specialized for decimal input/output.
function Int64HiLoToString(hi,lo){
hi>>>=0;lo>>>=0;
var sign="";
if(hi&0x80000000){
sign="-";
lo=(0x100000000-lo)>>>0;
hi=0xffffffff-hi+ +(lo===0);
}
var dhi=~~(hi/0x5af4),dhirem=hi%0x5af4;
var dlo=dhirem*0x100000000+dhi*0xef85c000+lo;
dhi += ~~(dlo/0x5af3107a4000);
dlo%=0x5af3107a4000;
var slo=""+dlo;
if(dhi){
slo="000000000000000000".slice(0,14-slo.length)+dlo;
return sign+dhi+slo;
}else{
return sign+slo;
}
}
Most likely this is what you needed.
I adapted the base conversion code from https://codegolf.stackexchange.com/questions/1620/arbitrary-base-conversion. Mistakes are mine, clevernesses are theirs.
I also had to add a bunch of code to deal with negative numbers (twos complement).
This code is ecmascript5, and will need slight reworking to work in older browsers.
function convert(hi, lo) {
function invertBit(bit) {
return bit == "0" ? "1" : "0";
}
function binaryInvert(binaryString) {
return binaryString.split("").map(invertBit).join("");
}
function binaryIncrement(binaryString) {
var idx = binaryString.lastIndexOf("0");
return binaryString.substring(0, idx) + "1" + binaryInvert(binaryString.substring(idx + 1));
}
function binaryDecrement(binaryString) {
var idx = binaryString.lastIndexOf("1");
return binaryString.substring(0, idx) + binaryInvert(binaryString.substring(idx));
}
function binaryAbs(binaryString) {
if (binaryString[0] === "1") {
return invertBits(binaryDecrement(binaryString));
}
return binaryString;
}
function to32Bits(val) {
var binaryString = val.toString(2);
if (binaryString[0] === "-") {
binaryString = Array(33 - (binaryString.length - 1)).join("1") + binaryInvert(binaryString.substr(1));
return binaryIncrement(binaryString);
}
return Array(33 - binaryString.length).join("0") + binaryString;
}
var fullBinaryNumber = to32Bits(hi) + to32Bits(lo);
var isNegative = fullBinaryNumber[0] === "1";
fullBinaryNumber = binaryAbs(fullBinaryNumber);
var result = "";
while (fullBinaryNumber.length > 0) {
var remainingToConvert = "", resultDigit = 0;
for (var position = 0; position < fullBinaryNumber.length; ++position) {
var currentValue = Number(fullBinaryNumber[position]) + resultDigit * 2;
var remainingDigitToConvert = Math.floor(currentValue / 10);
resultDigit = currentValue % 10;
if (remainingToConvert.length || remainingDigitToConvert) {
remainingToConvert += remainingDigitToConvert;
}
}
fullBinaryNumber = remainingToConvert;
result = resultDigit + result;
}
return (isNegative?"-":"") + result;
}
Examples:
> // largest negative number -2^63 (just the most significant bit set)
> convert(1 << 31, 0)
'-9223372036854775808'
> // largest positive number
> convert(0x7fffffff, 0xffffffff)
'9223372036854775807'
> // -1 is all bits set.
> convert(0xffffffff, 0xffffffff)
'-1'
According to JavaScript can't handle 64-bit integers, can it?, native numbers in Javascript have 53 bits of mantissa, so JS can't deal with 64 bits integers unless using specialized libraries.
Whatever the datatype and implementation limits, I assume you want to compute the Two's complement of the initial 64 bits unsigned number, to convert it from the [0 ... 2^64-1] range into the [-2^63 ... 2^63-1] range.
high is presumably the initial unsigned 64 bits number divided by 2^32, and low is the remainder.
The conversion to a signed 64 bits should go like this:
if high>=2^63 then
s64 = -(2^64-(high*2^32+low))
else
s64 = high*2^32+low;
In a PostgreSQL function, this can be done using the exact-precision numeric type to avoid overflows in intermediate multiplications, and downcast the final result to bigint (signed 64 bits):
create function concat64(bigint, bigint) returns bigint
as $$
select (case when $1>=2147483648
then -(18446744073709551616::numeric-($1*4294967296::numeric+$2))
else $1*4294967296::numeric+$2 end)::bigint;
$$ language sql;
The input arguments have to be bigint (64 bits) because postgres doesn't have unsigned types.
They're assumed to be in the [0..4294967296] range and the output should be in the [-9223372036854775808..9223372036854775807] range.
What I would like to have is the almost opposite of Number.prototype.toPrecision(), meaning that when i have number, how many decimals does it have? E.g.
(12.3456).getDecimals() // 4
For anyone wondering how to do this faster (without converting to string), here's a solution:
function precision(a) {
var e = 1;
while (Math.round(a * e) / e !== a) e *= 10;
return Math.log(e) / Math.LN10;
}
Edit: a more complete solution with edge cases covered:
function precision(a) {
if (!isFinite(a)) return 0;
var e = 1, p = 0;
while (Math.round(a * e) / e !== a) { e *= 10; p++; }
return p;
}
One possible solution (depends on the application):
var precision = (12.3456 + "").split(".")[1].length;
If by "precision" you mean "decimal places", then that's impossible because floats are binary. They don't have decimal places, and most values that have a small number of decimal places have recurring digits in binary, and when they're translated back to decimal that doesn't necessarily yield the original decimal number.
Any code that works with the "decimal places" of a float is liable to produce unexpected results on some numbers.
There is no native function to determine the number of decimals. What you can do is convert the number to string and then count the offset off the decimal delimiter .:
Number.prototype.getPrecision = function() {
var s = this + "",
d = s.indexOf('.') + 1;
return !d ? 0 : s.length - d;
};
(123).getPrecision() === 0;
(123.0).getPrecision() === 0;
(123.12345).getPrecision() === 5;
(1e3).getPrecision() === 0;
(1e-3).getPrecision() === 3;
But it's in the nature of floats to fool you. 1 may just as well be represented by 0.00000000989 or something. I'm not sure how well the above actually performs in real life applications.
Basing on #blackpla9ue comment and considering numbers exponential format:
function getPrecision (num) {
var numAsStr = num.toFixed(10); //number can be presented in exponential format, avoid it
numAsStr = numAsStr.replace(/0+$/g, '');
var precision = String(numAsStr).replace('.', '').length - num.toFixed().length;
return precision;
}
getPrecision(12.3456); //4
getPrecision(120.30003300000); //6, trailing zeros are truncated
getPrecision(15); //0
getPrecision(120.000)) //0
getPrecision(0.0000005); //7
getPrecision(-0.01)) //2
Try the following
function countDecimalPlaces(number) {
var str = "" + number;
var index = str.indexOf('.');
if (index >= 0) {
return str.length - index - 1;
} else {
return 0;
}
}
Based on #boolean_Type's method of handling exponents, but avoiding the regex:
function getPrecision (value) {
if (!isFinite(value)) { return 0; }
const [int, float = ''] = Number(value).toFixed(12).split('.');
let precision = float.length;
while (float[precision - 1] === '0' && precision >= 0) precision--;
return precision;
}
Here are a couple of examples, one that uses a library (BigNumber.js), and another that doesn't use a library. Assume you want to check that a given input number (inputNumber) has an amount of decimal places that is less than or equal to a maximum amount of decimal places (tokenDecimals).
With BigNumber.js
import BigNumber from 'bignumber.js'; // ES6
// const BigNumber = require('bignumber.js').default; // CommonJS
const tokenDecimals = 18;
const inputNumber = 0.000000000000000001;
// Convert to BigNumber
const inputNumberBn = new BigNumber(inputNumber);
// BigNumber.js API Docs: http://mikemcl.github.io/bignumber.js/#dp
console.log(`Invalid?: ${inputNumberBn.dp() > tokenDecimals}`);
Without BigNumber.js
function getPrecision(numberAsString) {
var n = numberAsString.toString().split('.');
return n.length > 1
? n[1].length
: 0;
}
const tokenDecimals = 18;
const inputNumber = 0.000000000000000001;
// Conversion of number to string returns scientific conversion
// So obtain the decimal places from the scientific notation value
const inputNumberDecimalPlaces = inputNumber.toString().split('-')[1];
// Use `toFixed` to convert the number to a string without it being
// in scientific notation and with the correct number decimal places
const inputNumberAsString = inputNumber.toFixed(inputNumberDecimalPlaces);
// Check if inputNumber is invalid due to having more decimal places
// than the permitted decimal places of the token
console.log(`Invalid?: ${getPrecision(inputNumberAsString) > tokenDecimals}`);
Assuming number is valid.
let number = 0.999;
let noOfPlaces = number.includes(".") //includes or contains
? number.toString().split(".").pop().length
: 0;
5622890.31 ops/s (91.58% slower):
function precision (n) {
return (n.toString().split('.')[1] || '').length
}
precision(1.0123456789)
33004904.53 ops/s (50.58% slower):
function precision (n) {
let e = 1
let p = 0
while(Math.round(n * e) / e !== n) {
e *= 10
p++
}
return p
}
precision(1.0123456789)
62610550.04 ops/s (6.25% slower):
function precision (n) {
let cur = n
let p = 0
while(!Number.isInteger(cur)) {
cur *= 10
p++
}
return p
}
precision(1.0123456789)
66786361.47 ops/s (fastest):
function precision (n) {
let cur = n
let p = 0
while(Math.floor(cur) !== cur) {
cur *= 10
p++
}
return p
}
precision(1.0123456789)
Here is a simple solution
First of all, if you pass a simple float value as 12.1234 then most of the below/above logics may work but if you pass a value as 12.12340, then it may exclude a count of 0. For e.g, if the value is 12.12340 then it may give you a result of 4 instead of 5. As per your problem statement, if you ask javascript to split and count your float value into 2 integers then it won't include trailing 0s of it.
Let's satisfy our requirement here with a trick ;)
In the below function you need to pass a value in string format and it will do your work
function getPrecision(value){
a = value.toString()
console.log('a ->',a)
b = a.split('.')
console.log('b->',b)
return b[1].length
getPrecision('12.12340') // Call a function
For an example, run the below logic
value = '12.12340'
a = value.toString()
b = a.split('.')
console.log('count of trailing decimals->',b[1].length)
That's it! It will give you the exact count for normal float values as well as the float values with trailing 0s!
Thank you!
This answer adds to Mourner's accepted solution by making the function more robust. As noted by many, floating point precision makes such a function unreliable. For example, precision(0.1+0.2) yields 17 rather than 1 (this might be computer specific, but for this example see https://jsfiddle.net/s0v17jby/5/).
IMHO, there are two ways around this: 1. either properly define a decimal type, using e.g. https://github.com/MikeMcl/decimal.js/, or 2. define an acceptable precision level which is both OK for your use case and not a problem for the js Number representation (8 bytes can safely represent a total of 16 digits AFAICT). For the latter workaround, one can write a more robust variant of the proposed function:
const MAX_DECIMAL_PRECISION = 9; /* must be <= 15 */
const maxDecimalPrecisionFloat = 10**MAX_DECIMAL_PRECISION;
function precisionRobust(a) {
if (!isFinite(a)) return 0;
var e = 1, p = 0;
while ( ++p<=MAX_DECIMAL_PRECISION && Math.round( ( Math.round(a * e) / e - a ) * maxDecimalPrecisionFloat ) !== 0) e *= 10;
return p-1;
}
In the above example, the maximum precision of 9 means this accepts up to 6 digits before the decimal point and 9 after (so this would work for numbers less than one million and with a maximum of 9 decimal points). If your use-case numbers are smaller then you can choose to make this precision even greater (but with a maximum of 15). It turns out that, for calculating precision, this function seems to do OK on larger numbers as well (though that would no longer be the case if we were, say, adding two rounded numbers within the precisionRobust function).
Finally, since we now know the maximum useable precision, we can further avoid infinite loops (which I have not been able to replicate but which still seem to cause problems for some).