I want to know about what exactly is being done with variable a in below function :
function c(a) {
var b = new Date;
return Math.round(b.getTime() / 1e3 + (a ? a : 0))
}
Code that needs clarity:
(a?a:0)
Just want to know what is the logic of highlighted text.
If a is undefined or false or null or NaN it will return 0 else it will return value of a
Lets assume
var someVar = 23;
function c(a) {
return a ? a : 0; //Also true for negative values
}
c(someVar); //will return 23
And
var someVar = -22;
c(someVar); //will return -22
And
var someVar = false;
c(someVar); //will return 0
And
var someVar; //someVar is undefined
c(someVar); //will return 0
You probably wanted to use bold style, aren't you?
The statement x > 5 ? true : false is a shortened version of an if...else statement. You put the statement before the "?". The ":" separates the if and else part. If the statement is true, then the part before the : fires, if false, then the one after it. Because javascript likes to convert anything into booleans, the statement you have is the same as if (a > 0) { b = a } else { b = 0}
If you want to know more about these statements, search for ternary operators
these two lines
var b = new Date;
return Math.round(b.getTime() / 1e3 + (a ? a : 0))
can also be written as (for better readability)
var b = new Date;
var c = b.getTime() / 1e3 ; // 1e3 is 1000, so c is basically number of seconds since 1970
if ( !a ) //if a is either undefined, null or false
{
a = 0;
}
return Math.round(c+a); //now c+a is adding these seconds to the paramter you have passed
So this function is basically passing the number of seconds since 1970 to the value you are passing.
the "? : " is known as a ternary operator. It is a shortcut for an if else. For example, var b = a ? a : 0 is equivalent to:
var b;
if(a){
b = a;
}else{
b = 0;
}
Also, for the sake of clarity, your code misses () and ;. Here is a correct version:
function c(a) {
var b = new Date();
return Math.round(b.getTime() / 1e3 + (a ? a : 0));
}
Have a look at this question for further explanation.
if(a){
return a;
}
else{
return 0;
}
Related
I have my js code for homework here. I have an if statement that should return -1 in console when the input is not a number but instead of returning -1 it returns NaN. Can anybody help me with this?
function calculateFoodOrder(numAnimals, avgFood) {
// IMPLEMENT THIS FUNCTION!
var total = avgFood*numAnimals;
if ((Number(numAnimals || avgFood) < 0) && (isNaN(numAnimals || avgFood))) {
console.log(-1);
} else {
return total
}
}
calculateFoodOrder()
Default values in your function are 'undefined'. That's why javascript is showing NaN. Define default values to avoid this. For example:
function calculateFoodOrder(numAnimals = 0, avgFood = 0) {
var total = avgFood*numAnimals;
if ((Number(numAnimals || avgFood) < 0) && (isNaN(numAnimals || avgFood))) {
console.log(-1);
} else {
return total;
}
}
If I understand your task correctly, you merely have to check both parameters for not being NaN using isNaN.
Your sample code does not work because any comparison with NaN returns false and thus Number(numAnimals || avgFood) < 0 is never going to be true. Check here for details on NaN.
function calculateFoodOrder(numAnimals, avgFood){
//REM: Default result
let tResult = -1;
//REM: Check if both paramters are not NaN
//REM: Be aware that whitespaces and empty strings validate to zero
if(
!isNaN(numAnimals) &&
!isNaN(avgFood)
){
tResult = Number(avgFood) * Number(numAnimals)
};
return tResult
};
//REM: Default test case
console.log(calculateFoodOrder());
;document.querySelector('button').onclick = function(){
console.log(
calculateFoodOrder(
document.getElementById('innumAnimals').value,
document.getElementById('inavgFood').value
)
)
};
<input type = 'text' value = '' id = 'innumAnimals' placeholder = 'numAnimals' />
<input type = 'text' value = '' id = 'inavgFood' placeholder = 'avgFood' />
<button>do</button>
Please do not confuse yourself with caveats if you are a beginner. You should assign to itself after converting to number, then judge if it's a NaN value or less then zero just like this,
function calculateFoodOrder(numAnimals, avgFood) {
// IMPLEMENT THIS FUNCTION!
numAnimals = Number(numAnimals);
avgFood = Number(avgFood);
var total = avgFood * numAnimals;
if (isNaN(numAnimals) || isNaN(avgFood) || numAnimals < 0 || avgFood < 0 ) {
console.log(-1);
} else {
return total
}
}
calculateFoodOrder()
I'm reading a book "Eloquent JavaScript" by Marijn Haverbeke and it says:
"The rules for converting strings and numbers to Boolean values state that 0, NaN, and the empty string ( "" ) count as false, while all the other values count as true."
Would be very nice if someone explains me what did the author mean by saying that NaN counts as false according to the rules of converting?
As I can see it now:
0 == false; // true
"" == false; // true
NaN == false; // false
0 / 0 == false; // false
Yes I know that "NaN is not equal to anything, even to the other NaN", but I just wonder what does the book want me to know?
Basically, if a variable is assigned to NaN, it will evaluate to false if used in a conditional statement. For example:
var b = NaN;
if(b){//code will never enter this block
console.log('the impossible has occurred!')
}
This is true as well if you get invalid input, for example:
var input = "abcd"
var num = parseFloat(input);
if(num){
console.log('the impossible has occurred!')
}
var a = NaN;
var b = null;
var c;
var d = false;
var e = 0;
document.write(a + " " + b + " " + c + " " + d + " "+ e + "<br>");
if(a || b || c || d || e){
document.write("Code won't enter here");
}
if(a || b || c || d || e || true){
document.write("Code finally enters");
}
Reference: link
Other answers are correct, but the significant thing here is that a conversion to a bool takes place in the case that it's used in a condition.
That's why:
NaN === false // false
Yet:
if(NaN) // false because it first does a boolean conversion
// which is the rule you see described
As a side note, NaN == false as used in the question (note == vs ===) actually does a type conversion from false to 0 per the == operator. It's beyond the scope of this question, but the difference in operators is well documented elsewhere.
The book wants you to know that JavaScript evaluates NaN to false.
var notANumber = 500 / 0;
if(notANumber) {
// this code block does not run
}
else {
// this code block runs
}
I have a string that contains Boolean logic something like:
var test = "(true)&&(false)&&!(true||true)"
What is a good way to evaluate this string in JavaScript to get the boolean value of false in this case
I know we could use eval() or new Function().. - but is that a safe approach?
I am guessing the other option would be to write a custom parser. Being a fairly new person to JS, would that be a lot of effort? I could not find any examples of parsers for Boolean logic expressions
Any other alternatives?
As long as you can guarantee it to be safe, I think you could use eval.
Maybe by treating it before doing an eval?
var test = "(true)&&(false)&&!(true||true)"
var safe = test.replace(/true/ig, "1").replace(/false/ig, "0");
var match = safe.match(/[0-9&!|()]*/ig);
if(match) {
var result = !!eval(match[0]);
}
Javascript has a ternary operator you could use:
var i = result ? 1 : 0;
Here, result is Boolean value either True or False.
So, Your question will be something like that after this operation.
(1)&(0)&!(1||1)
I hope you can better evaluate now this Boolean logic.
you can use eval,
Eg: eval("(true)&&(false)&&!(true||true)");
Try this code
function processExpression(expr)
{
while (expr.indexOf("(" ) != -1 )
{
expr = expr.replace(/\([\w|]+\)/g, function(matched){ return processBrace(matched)});
}
return expr = processBrace( "(" + expr + ")" );
}
function processBrace(str)
{
return str.substring(1).slice(0,-1).split(/(?=&|\|)/).map(function(value,index,arr){
if ( index != 0 && index%2 == 0 ) { return arr[index-1] + value } else if(index==0){return value;} else {return ""}
}).filter(function(val){return val.length > 0}).reduce(function(prev,current){
var first = Boolean(prev);
var operator = current.substring(0,2);
var operand = current.substring(2);
while ( operand.indexOf("!") != -1 )
{
var boolval = operand.match(/\w+/)[0] == "false"; //flip the value by comparing it with false
var negations = operand.match(/\W+/)[0];
operand = negations.substring(1) + boolval;
}
var second = operand == "true";
var output = operator == "&&" ? (first && second) : (first || second);
return output;
});
}
DEMO
function processExpression(expr)
{
while (expr.indexOf("(" ) != -1 )
{
expr = expr.replace(/\([\w|]+\)/g, function(matched){ return processBrace(matched)});
}
return expr = processBrace( "(" + expr + ")" );
}
function processBrace(str)
{
return str.substring(1).slice(0,-1).split(/(?=&|\|)/).map(function(value,index,arr){
if ( index != 0 && index%2 == 0 ) { return arr[index-1] + value } else if(index==0){return value;} else {return ""}
}).filter(function(val){return val.length > 0}).reduce(function(prev,current){
var first = Boolean(prev);
var operator = current.substring(0,2);
var operand = current.substring(2);
while ( operand.indexOf("!") != -1 )
{
var boolval = operand.match(/\w+/)[0] == "false"; //flip the value by comparing it with false
var negations = operand.match(/\W+/)[0];
operand = negations.substring(1) + boolval;
}
var second = operand == "true";
var output = operator == "&&" ? (first && second) : (first || second);
return output;
});
}
var example1 = "(true)&&(false)&&!(true||true)";
document.body.innerHTML += example1 + " -- " + processExpression(example1);
Try using "".match() in ternary operator condition
"(true)&&(true)&&!(true||true)".match(/false/ig)?false:true
Can anyone point me to some code to determine if a number in JavaScript is even or odd?
Use the below code:
function isOdd(num) { return num % 2;}
console.log("1 is " + isOdd(1));
console.log("2 is " + isOdd(2));
console.log("3 is " + isOdd(3));
console.log("4 is " + isOdd(4));
1 represents an odd number, while 0 represents an even number.
Use the bitwise AND operator.
function oddOrEven(x) {
return ( x & 1 ) ? "odd" : "even";
}
function checkNumber(argNumber) {
document.getElementById("result").innerHTML = "Number " + argNumber + " is " + oddOrEven(argNumber);
}
checkNumber(17);
<div id="result" style="font-size:150%;text-shadow: 1px 1px 2px #CE5937;" ></div>
If you don't want a string return value, but rather a boolean one, use this:
var isOdd = function(x) { return x & 1; };
var isEven = function(x) { return !( x & 1 ); };
You could do something like this:
function isEven(value){
if (value%2 == 0)
return true;
else
return false;
}
function isEven(x) { return (x%2)==0; }
function isOdd(x) { return !isEven(x); }
Do I have to make an array really large that has a lot of even numbers
No. Use modulus (%). It gives you the remainder of the two numbers you are dividing.
Ex. 2 % 2 = 0 because 2/2 = 1 with 0 remainder.
Ex2. 3 % 2 = 1 because 3/2 = 1 with 1 remainder.
Ex3. -7 % 2 = -1 because -7/2 = -3 with -1 remainder.
This means if you mod any number x by 2, you get either 0 or 1 or -1. 0 would mean it's even. Anything else would mean it's odd.
This can be solved with a small snippet of code:
function isEven(value) {
return !(value % 2)
}
Hope this helps :)
In ES6:
const isOdd = num => num % 2 == 1;
Like many languages, Javascript has a modulus operator %, that finds the remainder of division. If there is no remainder after division by 2, a number is even:
// this expression is true if "number" is even, false otherwise
(number % 2 == 0)
Similarly, if there is a remainder of 1 after division by 2, a number is odd:
// this expression is true if "number" is odd, false otherwise
(number % 2 == 1)
This is a very common idiom for testing for even integers.
With bitwise, codegolfing:
var isEven=n=>(n&1)?"odd":"even";
Use my extensions :
Number.prototype.isEven=function(){
return this % 2===0;
};
Number.prototype.isOdd=function(){
return !this.isEven();
}
then
var a=5;
a.isEven();
==False
a.isOdd();
==True
if you are not sure if it is a Number , test it by the following branching :
if(a.isOdd){
a.isOdd();
}
UPDATE :
if you would not use variable :
(5).isOdd()
Performance :
It turns out that Procedural paradigm is better than OOP paradigm .
By the way , i performed profiling in this FIDDLE . However , OOP way is still prettiest .
A simple function you can pass around. Uses the modulo operator %:
var is_even = function(x) {
return !(x % 2);
}
is_even(3)
false
is_even(6)
true
if (X % 2 === 0){
} else {
}
Replace X with your number (can come from a variable). The If statement runs when the number is even, the Else when it is odd.
If you just want to know if any given number is odd:
if (X % 2 !== 0){
}
Again, replace X with a number or variable.
<script>
function even_odd(){
var num = document.getElementById('number').value;
if ( num % 2){
document.getElementById('result').innerHTML = "Entered Number is Odd";
}
else{
document.getElementById('result').innerHTML = "Entered Number is Even";
}
}
</script>
</head>
<body>
<center>
<div id="error"></div>
<center>
<h2> Find Given Number is Even or Odd </h2>
<p>Enter a value</p>
<input type="text" id="number" />
<button onclick="even_odd();">Check</button><br />
<div id="result"><b></b></div>
</center>
</center>
</body>
Many people misunderstand the meaning of odd
isOdd("str") should be false.
Only an integer can be odd.
isOdd(1.223) and isOdd(-1.223) should be false.
A float is not an integer.
isOdd(0) should be false.
Zero is an even integer (https://en.wikipedia.org/wiki/Parity_of_zero).
isOdd(-1) should be true.
It's an odd integer.
Solution
function isOdd(n) {
// Must be a number
if (isNaN(n)) {
return false;
}
// Number must not be a float
if ((n % 1) !== 0) {
return false;
}
// Integer must not be equal to zero
if (n === 0) {
return false;
}
// Integer must be odd
if ((n % 2) !== 0) {
return true;
}
return false;
}
JS Fiddle (if needed): https://jsfiddle.net/9dzdv593/8/
1-liner
Javascript 1-liner solution. For those who don't care about readability.
const isOdd = n => !(isNaN(n) && ((n % 1) !== 0) && (n === 0)) && ((n % 2) !== 0) ? true : false;
You can use a for statement and a conditional to determine if a number or series of numbers is odd:
for (var i=1; i<=5; i++)
if (i%2 !== 0) {
console.log(i)
}
This will print every odd number between 1 and 5.
Just executed this one in Adobe Dreamweaver..it works perfectly.
i used if (isNaN(mynmb))
to check if the given Value is a number or not,
and i also used Math.abs(mynmb%2) to convert negative number to positive and calculate
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
</head>
<body bgcolor = "#FFFFCC">
<h3 align ="center"> ODD OR EVEN </h3><table cellspacing = "2" cellpadding = "5" bgcolor="palegreen">
<form name = formtwo>
<td align = "center">
<center><BR />Enter a number:
<input type=text id="enter" name=enter maxlength="10" />
<input type=button name = b3 value = "Click Here" onClick = compute() />
<b>is<b>
<input type=text id="outtxt" name=output size="5" value="" disabled /> </b></b></center><b><b>
<BR /><BR />
</b></b></td></form>
</table>
<script type='text/javascript'>
function compute()
{
var enter = document.getElementById("enter");
var outtxt = document.getElementById("outtxt");
var mynmb = enter.value;
if (isNaN(mynmb))
{
outtxt.value = "error !!!";
alert( 'please enter a valid number');
enter.focus();
return;
}
else
{
if ( mynmb%2 == 0 ) { outtxt.value = "Even"; }
if ( Math.abs(mynmb%2) == 1 ) { outtxt.value = "Odd"; }
}
}
</script>
</body>
</html>
When you need to test if some variable is odd, you should first test if it is integer. Also, notice that when you calculate remainder on negative number, the result will be negative (-3 % 2 === -1).
function isOdd(value) {
return typeof value === "number" && // value should be a number
isFinite(value) && // value should be finite
Math.floor(value) === value && // value should be integer
value % 2 !== 0; // value should not be even
}
If Number.isInteger is available, you may also simplify this code to:
function isOdd(value) {
return Number.isInteger(value) // value should be integer
value % 2 !== 0; // value should not be even
}
Note: here, we test value % 2 !== 0 instead of value % 2 === 1 is because of -3 % 2 === -1. If you don't want -1 pass this test, you may need to change this line.
Here are some test cases:
isOdd(); // false
isOdd("string"); // false
isOdd(Infinity); // false
isOdd(NaN); // false
isOdd(0); // false
isOdd(1.1); // false
isOdd("1"); // false
isOdd(1); // true
isOdd(-1); // true
Using % will help you to do this...
You can create couple of functions to do it for you... I prefer separte functions which are not attached to Number in Javascript like this which also checking if you passing number or not:
odd function:
var isOdd = function(num) {
return 'number'!==typeof num ? 'NaN' : !!(num % 2);
};
even function:
var isEven = function(num) {
return isOdd(num)==='NaN' ? isOdd(num) : !isOdd(num);
};
and call it like this:
isOdd(5); // true
isOdd(6); // false
isOdd(12); // false
isOdd(18); // false
isEven(18); // true
isEven('18'); // 'NaN'
isEven('17'); // 'NaN'
isOdd(null); // 'NaN'
isEven('100'); // true
A more functional approach in modern javascript:
const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")
const negate = f=> (...args)=> !f(...args)
const isOdd = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = negate(isOdd)
One liner in ES6 just because it's clean.
const isEven = (num) => num % 2 == 0;
Subtract 2 to it recursively until you reach either -1 or 0 (only works for positive integers obviously) :)
Every odd number when divided by two leaves remainder as 1 and every even number when divided by zero leaves a zero as remainder. Hence we can use this code
function checker(number) {
return number%2==0?even:odd;
}
How about this...
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);
This is what I did
//Array of numbers
var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10,32,23,643,67,5876,6345,34,3453];
//Array of even numbers
var evenNumbers = [];
//Array of odd numbers
var oddNumbers = [];
function classifyNumbers(arr){
//go through the numbers one by one
for(var i=0; i<=arr.length-1; i++){
if (arr[i] % 2 == 0 ){
//Push the number to the evenNumbers array
evenNumbers.push(arr[i]);
} else {
//Push the number to the oddNumbers array
oddNumbers.push(arr[i]);
}
}
}
classifyNumbers(numbers);
console.log('Even numbers: ' + evenNumbers);
console.log('Odd numbers: ' + oddNumbers);
For some reason I had to make sure the length of the array is less by one. When I don't do that, I get "undefined" in the last element of the oddNumbers array.
I'd implement this to return a boolean:
function isOdd (n) {
return !!(n % 2);
// or ((n % 2) !== 0).
}
It'll work on both unsigned and signed numbers. When the modulus return -1 or 1 it'll get translated to true.
Non-modulus solution:
var is_finite = isFinite;
var is_nan = isNaN;
function isOdd (discriminant) {
if (is_nan(discriminant) && !is_finite(discriminant)) {
return false;
}
// Unsigned numbers
if (discriminant >= 0) {
while (discriminant >= 1) discriminant -= 2;
// Signed numbers
} else {
if (discriminant === -1) return true;
while (discriminant <= -1) discriminant += 2;
}
return !!discriminant;
}
By using ternary operator, you we can find the odd even numbers:
var num = 2;
result = (num % 2 == 0) ? 'even' : 'odd'
console.log(result);
Another example using the filter() method:
let even = arr.filter(val => {
return val % 2 === 0;
});
// even = [2,4,6]
So many answers here but i just have to mention one point.
Normally it's best to use the modulo operator like % 2 but you can also use the bitwise operator like & 1. They both would yield the same outcome. However their precedences are different. Say if you need a piece of code like
i%2 === p ? n : -n
it's just fine but with the bitwise operator you have to do it like
(i&1) === p ? n : -n
So there is that.
this works for arrays:
function evenOrOdd(numbers) {
const evenNumbers = [];
const oddNumbers = [];
numbers.forEach(number => {
if (number % 2 === 0) {
evenNumbers.push(number);
} else {
oddNumbers.push(number);
}
});
console.log("Even: " + evenNumbers + "\nOdd: " + oddNumbers);
}
evenOrOdd([1, 4, 9, 21, 41, 92]);
this should log out:
4,92
1,9,21,41
for just a number:
function evenOrOdd(number) {
if (number % 2 === 0) {
return "even";
}
return "odd";
}
console.log(evenOrOdd(4));
this should output even to the console
A Method to know if the number is odd
let numbers = [11, 20, 2, 5, 17, 10];
let n = numbers.filter((ele) => ele % 2 != 0);
console.log(n);
I decided to create simple isEven and isOdd function with a very simple algorithm:
function isEven(n) {
n = Number(n);
return n === 0 || !!(n && !(n%2));
}
function isOdd(n) {
return isEven(Number(n) + 1);
}
That is OK if n is with certain parameters, but fails for many scenarios. So I set out to create robust functions that deliver correct results for as many scenarios as I could, so that only integers within the limits of javascript numbers are tested, everything else returns false (including + and - infinity). Note that zero is even.
// Returns true if:
//
// n is an integer that is evenly divisible by 2
//
// Zero (+/-0) is even
// Returns false if n is not an integer, not even or NaN
// Guard against empty string
(function (global) {
function basicTests(n) {
// Deal with empty string
if (n === '')
return false;
// Convert n to Number (may set to NaN)
n = Number(n);
// Deal with NaN
if (isNaN(n))
return false;
// Deal with infinity -
if (n === Number.NEGATIVE_INFINITY || n === Number.POSITIVE_INFINITY)
return false;
// Return n as a number
return n;
}
function isEven(n) {
// Do basic tests
if (basicTests(n) === false)
return false;
// Convert to Number and proceed
n = Number(n);
// Return true/false
return n === 0 || !!(n && !(n%2));
}
global.isEven = isEven;
// Returns true if n is an integer and (n+1) is even
// Returns false if n is not an integer or (n+1) is not even
// Empty string evaluates to zero so returns false (zero is even)
function isOdd(n) {
// Do basic tests
if (basicTests(n) === false)
return false;
// Return true/false
return n === 0 || !!(n && (n%2));
}
global.isOdd = isOdd;
}(this));
Can anyone see any issues with the above? Is there a better (i.e. more accurate, faster or more concise without being obfuscated) version?
There are various posts relating to other languages, but I can't seem to find a definitive version for ECMAScript.
Use modulus:
function isEven(n) {
return n % 2 == 0;
}
function isOdd(n) {
return Math.abs(n % 2) == 1;
}
You can check that any value in Javascript can be coerced to a number with:
Number.isFinite(parseFloat(n))
This check should preferably be done outside the isEven and isOdd functions, so you don't have to duplicate error handling in both functions.
I prefer using a bit test:
if(i & 1)
{
// ODD
}
else
{
// EVEN
}
This tests whether the first bit is on which signifies an odd number.
How about the following? I only tested this in IE, but it was quite happy to handle strings representing numbers of any length, actual numbers that were integers or floats, and both functions returned false when passed a boolean, undefined, null, an array or an object. (Up to you whether you want to ignore leading or trailing blanks when a string is passed in - I've assumed they are not ignored and cause both functions to return false.)
function isEven(n) {
return /^-?\d*[02468]$/.test(n);
}
function isOdd(n) {
return /^-?\d*[13579]$/.test(n);
}
Note: there are also negative numbers.
function isOddInteger(n)
{
return isInteger(n) && (n % 2 !== 0);
}
where
function isInteger(n)
{
return n === parseInt(n, 10);
}
Why not just do this:
function oddOrEven(num){
if(num % 2 == 0)
return "even";
return "odd";
}
oddOrEven(num);
To complete Robert Brisita's bit test .
if ( ~i & 1 ) {
// Even
}
var isOdd = x => Boolean(x % 2);
var isEven = x => !isOdd(x);
var isEven = function(number) {
// Your code goes here!
if (number % 2 == 0){
return(true);
}
else{
return(false);
}
};
A few
x % 2 == 0; // Check if even
!(x & 1); // bitmask the value with 1 then invert.
((x >> 1) << 1) == x; // divide value by 2 then multiply again and check against original value
~x&1; // flip the bits and bitmask
We just need one line of code for this!
Here a newer and alternative way to do this, using the new ES6 syntax for JS functions, and the one-line syntax for the if-else statement call:
const isEven = num => ((num % 2) == 0);
alert(isEven(8)); //true
alert(isEven(9)); //false
alert(isEven(-8)); //true
A simple modification/improvement of Steve Mayne answer!
function isEvenOrOdd(n){
if(n === parseFloat(n)){
return isNumber(n) && (n % 2 == 0);
}
return false;
}
Note: Returns false if invalid!
Different way:
var isEven = function(number) {
// Your code goes here!
if (((number/2) - Math.floor(number/2)) === 0) {return true;} else {return false;};
};
isEven(69)
Otherway using strings because why not
function isEven(__num){
return String(__num/2).indexOf('.') === -1;
}
if (testNum == 0);
else if (testNum % 2 == 0);
else if ((testNum % 2) != 0 );
Maybe this?
if(ourNumber % 2 !== 0)
var num = someNumber
isEven;
parseInt(num/2) === num/2 ? isEven = true : isEven = false;
for(var a=0; a<=20;a++){
if(a%2!==0){
console.log("Odd number "+a);
}
}
for(var b=0; b<=20;a++){
if(b%2===0){
console.log("Even number "+b);
}
}
Check if number is even in a line of code:
var iseven=(_)=>_%2==0
This one is more simple!
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);
To test whether or not you have a odd or even number, this also works.
const comapare = x => integer(checkNumber(x));
function checkNumber (x) {
if (x % 2 == 0) {
return true;
}
else if (x % 2 != 0) {
return false;
}
}
function integer (x) {
if (x) {
console.log('even');
}
else {
console.log('odd');
}
}
Using modern javascript style:
const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")
const isOdd = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = n=> isOdd(+n+1)
function isEven(n) {return parseInt(n)%2===0?true:parseInt(n)===0?true:false}
when 0/even wanted but
isEven(0) //true
isEven(1) //false
isEven(2) //true
isEven(142856) //true
isEven(142856.142857)//true
isEven(142857.1457)//false
if (i % 2) {
return odd numbers
}
if (i % 2 - 1) {
return even numbers
}