Decimal number to binary number conversion in JavaScript

Question

I made this decimal to binary conversion as an exercise for myself and because it is obviously often asked during job interviews / tests.

Any hints concerning possible improvements welcomed.

// -- START : Testing the function --------
var display = document.querySelector('div');
var result = '';
var n = 255;
var maxSpace = n.toString().length;
// Assigning wrong parameter.
console.log('Result empty: ' + getBinaryNumber());
console.log('Result string: ' + getBinaryNumber('abc'));
// Helper function: Avoiding "stairs".
function getSpaces(count, someChar) {
 
 if (!count) return '';
 
 var ret = '';
 
 for (var i = 0; i < count; i++) {
 ret += someChar;
 }
 
 return ret;
}
var spaces = getSpaces(maxSpace, '0');
for (var i = 0; i <= n; i++) {
 result += 
 (spaces + i).slice(- maxSpace) + 
 ' : ' + getBinaryNumber(i, ' ') + '<br />';
}
display.innerHTML = result;
// -- END : Testing the function --------
// Converts a decimal number to a binary number.
// -- Parameter --------------------------
// Number - The decimal number to convert.
// -- Return -----------------------------
// String - The number in it's binary
// represention.
// On error: 
// Returns an empty string.
function getBinaryNumber (decimalNumber, separator) {
 // Parameter check
 if (decimalNumber === 0) return 0;
 
 if (!decimalNumber) return '';
 
 if (isNaN(decimalNumber)) return '';
 
 separator = separator || '';
 // Local variables
 var ret = [];
 var bit = decimalNumber % 2; 
 
 ret.push(bit);
 
 decimalNumber = ~~(decimalNumber / 2);
 
 while (decimalNumber > 1) {
 bit = decimalNumber % 2; 
 ret.push(bit);
 
 decimalNumber = ~~(decimalNumber / 2);
 }
 
 if (decimalNumber > 0) ret.push(decimalNumber);
 
 return ret.reverse().join(separator);
}

<div></div>

• Julian: Hint: PHP can achieve such thing by using base_convert. You would need less code in comparison with js. php.net/manual/en/function.base-convert.php

Answer 1

A general feeling after taking a glance at your code is that you do this:

if (decimalNumber > 0) ret.push(decimalNumber);

i.e. that you put in the same line of code, both the if statement and the if body.

I would expect to see (with brackets maybe):

if (decimalNumber > 0)
 ret.push(decimalNumber);

That might be of course a personal taste, but I thought it would be nice for you to think twice about it.

---

Also:

// Local variables
var ret = [];
var bit = decimalNumber % 2; 

everybody knows that ret and bit are local variables, so, think twice2 for removing that comment.

• michael.zech: Had the feeling that I need some kind of separator when opening another "section" after the parameter-checks. But yes you are right. It's a bit stupid within there.

Answer 2

Comment says, that function will return a string, and then returns a number.

if (decimalNumber === 0) return 0

Best improvement would be using a native method

function getBinaryNumber(num, sep) {
 if (typeof num != "number") num = parseFloat(num)
 return sep ? num.toString(2).split("").join(sep) : num.toString(2)
}

I would remove separator parameter, as it makes no sense to combine this functionality.

function getBinaryNumber(num) {
 return (typeof num == "number" ? num : parseFloat(num)).toString(2)
}

• michael.zech: Thanks a lot. You've given me lots of useful hints. :) But concerning the isNaN-check you are wrong. It shall intercept the case: Something which can't be converted to a number (e.g. 'a#bc') has been given as parameter. Without that check 'bit' gets the value NaN after applying the mod-operator the first time. This would become the return value later.
• Lauri: My bad about isNaN, removed this from answer