Develop Reference

Decode a byte array to a signed integer up to 64 bit

I know that JavaScript can't precisely represent all 64 bit integer numbers. But it can precisely represent numbers larger than 32 bit. And that's what I need. With whatever precision JavaScript can give me.
I have a byte array of known length. It has 1, 2, 4, 8 or 16 bytes. And it can contain a signed or unsigned integer, I know which it is. The data is big-endian (network byte order).
How can I get the number value from that byte array?
There are simple solitions that completely fail on negative numbers. There's DataView that isn't of any help with more than 32 bits. I'm interested in a nice and simple and preferable efficient pure JavaScript solution to handle this. There doesn't seem to be any solution for this in the part of the web that's visible to me.
In case somebody wants to see wrong code, here is my version of positive numbers:
function readInt(array) {
var value = 0;
for (var i = array.length - 1; i >= 0; i--) {
value = (value * 256) + array[i];
}
return value;
}

This page explains a good and simple solution:
Add up all bits from the right end, up to the second most significant bit.
The most significant bit is not added as 2^i but as -2^i.
My code works in a larger scope that has an array and a pos to read from.
function readInt(size) {
var value = 0;
var first = true;
while (size--) {
if (first) {
let byte = array[pos++];
value += byte & 0x7f;
if (byte & 0x80) {
value -= 0x80; // Treat most-significant bit as -2^i instead of 2^i
}
first = false;
}
else {
value *= 256;
value += array[pos++];
}
}
return value;
}
The raw bytes are provided in array (a Uint8Array) and pos is the next index to read. This function starts to read at the current pos and advances pos as it reads one of the size bytes.

Counting numbers after decimal point in JavaScript

I have a problem in JavaScript. Is it possible to check how many numbers are after the decimal point? I tried to do it using a.toString().split(".")[1]), but if there is no decimal point in the number, there is an error. What should I do if I want the system to do nothing if there is no decimal point?

You're on the right track. You can also .includes('.') to test if it contains a decimal along with .length to return the length of the decimal portion.
function decimalCount (number) {
// Convert to String
const numberAsString = number.toString();
// String Contains Decimal
if (numberAsString.includes('.')) {
return numberAsString.split('.')[1].length;
}
// String Does Not Contain Decimal
return 0;
}
console.log(decimalCount(1.123456789)) // 9
console.log(decimalCount(123456789)) // 0

Convert to a string, split on “.”, then when there is no “.” to split on, assume it’s empty string '' (the part you’re missing), then get said string’s length:
function numDigitsAfterDecimal(x) {
var afterDecimalStr = x.toString().split('.')[1] || ''
return afterDecimalStr.length
}
console.log(numDigitsAfterDecimal(1.23456))
console.log(numDigitsAfterDecimal(0))

You could check if no dot is available, then return zero, otherwise return the delta of the lenght and index with an adjustment.
function getDigits(v) {
var s = v.toString(),
i = s.indexOf('.') + 1;
return i && s.length - i;
}
console.log(getDigits(0));
console.log(getDigits(0.002));
console.log(getDigits(7.654321));
console.log(getDigits(1234567890.654321));

The condition you need is:
number.split('.')[1].length
It checks if there are any numbers after the dot which separates the number from its decimal part.
I'm not sure if you are able to use split on numbers though. If not, parse it to a string.

You first need to convert the decimal number to string and then get the count of character after decimal point,
var a = 10.4578;
var str = a.toString();
if(str){
var val = str.split('.');
if(val && val.length == 2){
alert('Length of number after decimal point is ', val[1].length);
} else {
alert('Not a decimal number');
}
}
The output is 4

What does this line in the function do?

Ok, so I have this function, I just want to break down what exactly it is returning.
function componentToHex(c) {
var hex = c.toString(16);
return hex.length == 1 ? "0" + hex : hex;
}
Let's assume: c = 76
And therefore:
c.toString(16) = 4c
So what is the next bit doing?

It is returning the result of a Conditional (Ternary) Operator, which checks the length of the string hex and returns either 0 plus the string hex or the string hex.
In a long version it is this:
if (hex.length == 1) {
return "0" + hex;
} else {
return hex;
}

most likely it is used in color codes context and converting a color component (red, green or blue) to it's hexadecimal representation. Then, if the result is one digit, it prepends a zero in order to finally get, when all components have been converted a six character hex color code.

How can I check that a bit is set (without bitwise operation)?

Looking at the int 44 — I need Math.CEIL (log(2) 44) of binary places to represent 44.
(answer is 6 places)
6 places :
___ ___ ___ ___ ___ ___
32 16 8 4 2 1
But how can I check that (for example) the bit of 8 is checked or not ?
A simple solution will be do to :
((1<<3) & 44)>0 so this will check if the bit is set.
But please notice that behind the scenes the computer translates 44 to its binary representation and just check if bit is set via bitwise operation.
Another solution is just to build the binary myself via toString(2) or mod%2 in a loop
Question
Mathematically Via which formula, I can test if n'th bit is set ?
(I would prefer a non loop operation but pure single math phrase)

Divide by the value of the bit that you want to check
and test if the first bit is set (this can be tested with x mod 2 == 1)
Math expression:
floor(value/(2^bitPos)) mod 2 = 1
As JS function:
function isSet(value, bitPos) {
var result = Math.floor(value / Math.pow(2, bitPos)) % 2;
return result == 1;
}
Note: bitPos starts with 0 (bit representing the nr 1)

The 'bit' (actually any base) value of an indexed number index in a value val in base base can in general be calculated as
val = 1966;
index = 2;
base = 10;
alert (Math.floor(val/Math.pow(base,index)) % base);
result: 9
val = 44;
index = 3;
base = 2;
alert (Math.floor(val/Math.pow(base,index)) % base);
result: 1 (only 0 and 1 are possible here – the range will always be 0..base-1).
The combination of Math.floor (to coerce to an integer in Javascript) and Math.pow is kind of iffy here. Even in integer range, Math.pow may generate a floating point number slightly below the expected 'whole' number. Perhaps it is safer to always add a small constant:
alert (Math.floor(0.1+val/Math.pow(base,index)) % base);

You can simply check if the bit at the position is set to 1.
function isBitSet(no, index) {
var bin = no.toString(2);
// Convert to Binary
index = bin.length - index;
// Reverse the index, start from right to left
return bin[index] == 1;
}
isBitSet(44, 2); // Check if second bit is set from left
DEMO

Javascript: Comparing two float values [duplicate]

This question already has answers here:
Javascript float comparison
(2 answers)
Closed 6 months ago.
I have this JavaScript function:
Contrl.prototype.EvaluateStatement = function(acVal, cfVal) {
var cv = parseFloat(cfVal).toFixed(2);
var av = parseFloat(acVal).toFixed(2);
if( av < cv) // do some thing
}
When i compare float numbers av=7.00 and cv=12.00 the result of 7.00<12.00 is false!
Any ideas why?

toFixed returns a string, and you are comparing the two resulting strings. Lexically, the 1 in 12 comes before the 7 so 12 < 7.
I guess you want to compare something like:
(Math.round(parseFloat(acVal)*100)/100)
which rounds to two decimals

Compare float numbers with precision:
var precision = 0.001;
if (Math.abs(n1 - n2) <= precision) {
// equal
}
else {
// not equal
}
UPD:
Or, if one of the numbers is precise, compare precision with the relative error
var absoluteError = (Math.abs(nApprox - nExact)),
relativeError = absoluteError / nExact;
return (relativeError <= precision);

The Math.fround() function returns the nearest 32-bit single precision float representation of a Number.
And therefore is one of the best choices to compare 2 floats.
if (Math.fround(1.5) < Math.fround(1.6)) {
console.log('yes')
} else {
console.log('no')
}
>>> yes
// More examples:
console.log(Math.fround(0.9) < Math.fround(1)); >>> true
console.log(Math.fround(1.5) < Math.fround(1.6)); >>> true
console.log(Math.fround(0.005) < Math.fround(0.00006)); >>> false
console.log(Math.fround(0.00000000009) < Math.fround(0.0000000000000009)); >>> false

Comparing floats using short notation, also accepts floats as strings and integers:
var floatOne = 2, floatTwo = '1.456';
Math.floor(floatOne*100) > Math.floor(floatTwo*100)
(!) Note: Comparison happens using integers. What actually happens behind the scenes: 200 > 145
Extend 100 with zero's for more decimal precision. For example use 1000 for 3 decimals precision.
Test:
var floatOne = 2, floatTwo = '1.456';
console.log(Math.floor(floatOne*100), '>', Math.floor(floatTwo*100), '=', Math.floor(floatOne*100) > Math.floor(floatTwo*100));

Comparing of float values is tricky due to long "post dot" tail of the float value stored in the memory. The simplest (and in fact the best) way is: to multiply values, for reducing known amount of post dot digits to zero, and then round the value (to rid of the tail).
Obviously both compared values must be multiplied by the same rate.
F.i.: 1,234 * 1000 gives 1234 - which can be compared very easily. 5,67 can be multiplied by 100, as for reducing the float comparing problem in general, but then it couldn't be compared to the first value (1,234 vel 1234). So in this example it need to be multiplied by 1000.
Then the comparition code could look like (in meta code):
var v1 = 1.234;
var v2 = 5.67;
if (Math.round(v1*1000) < Math.round(v2*1000)) ....