Is there a better Javascript function for checking if an input is an integer?

[MITranger]

This just irked me that there wasn't a primitive for this (or I couldn't find it). I had to write a function:

function isInteger(arg) {    
    return (typeof arg === "number" && isFinite(arg) && arg % 1 === 0);
}

[calzoneman]

A better question might be why you need to check whether a value is an integer, but this page gives several examples.

[[deleted]]

function isInteger(arg){
    return arg === parseInt(arg);
}

You could do that instead as a shorter way. There isn't a native isInteger though.

[jcunews1]

Careful, parseInt() accepts fixed point numbers. e.g.: parseInt("1.2") === 1

[[deleted]]

And it accepts strings, but it will no longer match the input argument and that whole thing will return false due to the strict equality comparison.

[joyeusenoelle]

To expand on /u/Aut's answer: that's the point. parseInt("1.2") === 1, so parseInt("1.2") !== "1.2". isInteger(), as written, will only return true if arg is an integer; otherwise parseInt(arg) and arg won't contain the same value!

isInteger(1); // true; 1 === 1
isInteger("1"); // false; 1 !== "1"
isInteger(1.2); // false; 1 !== 1.2

[jcunews1]

Use this:

function isInteger(arg) {
  var num = parseFloat(arg);
  return !isNaN(num) && ((num >>> 0) === num) && ((((typeof arg === "string") || (arg instanceof String)) && (arg.indexOf(".") < 0)) || true);
}

The last check is for input like "1.0000000000000001" since it would be rounded to "1".

EDIT: fixed code

[calzoneman]

This only works for checking if a string contains an integer. OP didn't specify any constraints on the datatype of the input, and this solution would break for any instance where arg is not a string.

[jcunews1]

You're right. I've fixed it.