Convert integer MAC address to string in Javascript

The accepted answer did not pass my unit tests (convert mac address to a number then convert back and expect the result equals the input).

As @blueren said the accepted answer returns a reversed result.

My version is:

(mac_num).toString(16).padStart(12, '0').match(/../g)
                .reverse()
                .slice(0, 6)
                .reverse()
                .join(':')

My tests:

        expect(mac.toString(mac.toLong('b4:ae:2b:db:1c:3d'))).to.deep.equal('b4:ae:2b:db:1c:3d');
        expect(mac.toString(mac.toLong('00:00:00:db:1c:3d'))).to.deep.equal('00:00:00:db:1c:3d');
        expect(mac.toString(mac.toLong('b4:ae:2b:00:00:00'))).to.deep.equal('b4:ae:2b:00:00:00');
        expect(mac.toString(mac.toLong('00:00:00:00:00:00'))).to.deep.equal('00:00:00:00:00:00');
        expect(mac.toString(mac.toLong('10:00:00:00:00:00'))).to.deep.equal('10:00:00:00:00:00');
        expect(mac.toString(mac.toLong('00:00:01:00:00:00'))).to.deep.equal('00:00:01:00:00:00');
        expect(mac.toString(mac.toLong('00:00:00:00:00:01'))).to.deep.equal('00:00:00:00:00:01');

The following does the job,

var hexValue = parseInt('44873434449413').toString(16);

var macaddress = [];

for (var i=0; i < hexValue.length; i=i+2) {
    macaddress.push(hexValue.substr(i,2));    
}


console.log(macaddress.join(':'));

Output:
28:cf:e9:1e:c6:05

EDIT:

to take care of trailing 0's

str='3';
if (str.length < 12) { str = pad_after(str, 12, 0);}

var hexValue = parseInt(str).toString(16);

if (hexValue.length < 12) { hexValue = pad_before(hexValue, 12, 0);}

var macaddress = [];

for (var i=0; i < hexValue.length; i=i+2) {
    macaddress.push(hexValue.substr(i,2));    
}

console.log(macaddress.join(':'));


function pad_before(n, width, z) {
  z = z || '0';
  n = n + '';
  return n.length >= width ? n : new Array(width - n.length + 1).join(z) + n;
}

function pad_after(n, width, z) {
  z = z || '0';
  n = n + '';
  return n.length >= width ? n : n + new Array(width - n.length + 1).join(z);
}

Output:
00:45:d9:64:b8:00

A simple approach looks as follows:

var mac = 81952921372024;

mac.toString( 16 )             // "4a8926c44578"
        .match( /.{1,2}/g )    // ["4a", "89", "26", "c4", "45", "78"]
        .reverse()             // ["78", "45", "c4", "26", "89", "4a"]
        .join( ':' )           // "78:45:c4:26:89:4a"

> "78:45:c4:26:89:4a"

However I suggest putting additional 00 groups just for pathological cases when your integer is very short (i.e. mac = 150):

var mac = 81952921372024;

new Array( 6 ).join( '00' )    // '000000000000'
    .match( /../g )            // [ '00', '00', '00', '00', '00', '00' ]
    .concat( 
        mac.toString( 16 )     // "4a8926c44578"
           .match( /.{1,2}/g ) // ["4a", "89", "26", "c4", "45", "78"]
    )                          // ["00", "00", "00", "00", "00", "00", "4a", "89", "26", "c4", "45", "78"]
    .reverse()                 // ["78", "45", "c4", "26", "89", "4a", "00", "00", "00", "00", "00", "00", ]
    .slice( 0, 6 )             // ["78", "45", "c4", "26", "89", "4a" ]
    .join( ':' )               // "78:45:c4:26:89:4a"

> "78:45:c4:26:89:4a"