Question

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Working with hex strings and hex values more easily in Javascript

rated 0 times [ 4] [ 6] / answers: 1 / hits: 83644 / 12 Years ago, wed, june 13, 2012, 12:00:00

I have some code which takes strings representing hexadecimal numbers - hex colors, actually - and adds them. For example, adding aaaaaa and 010101 gives the output ababab.

However, my method seems unnecessarily long and complicated:

var hexValue = aaaaaa;

hexValue = 0x + hexValue;

hexValue = parseInt(hexValue, 16);

hexValue = hexValue + 0x010101;

hexValue = hexValue.toString(16);

document.write(hexValue); // outputs 'ababab'

The hex value is still a string after concatenating 0x, so then I have to change it to a number, then I can add, and then I have to change it back into hex format! There are even more steps if the number I'm adding to it is a hexadecimal string to begin with, or if you take into consideration that I am removing the # from the hex color before all this starts.

Surely there's a simpler way to do such simple hexadecimal calculations! And just to be clear, I don't mean just putting it all on one line like (parseInt("0x"+"aaaaaa",16)+0x010101).toString(16) or using shorthand - I mean actually doing less operations.

Is there some way to get javascript to stop using decimal for all of its mathematical operations and use hex instead? Or is there some other method of making JS work with hex more easily?

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havenbilliec

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3 Years ago

answered 12 Years ago kieraelsies · Accepted Answer

No, there is no way to tell the JavaScript language to use hex integer format instead of decimal by default. Your code is about as concise as it gets but note that you do not need to prepend the "0x" base indicator when you use "parseInt" with a base.

Here is how I would approach your problem:

function addHexColor(c1, c2) {

  var hexStr = (parseInt(c1, 16) + parseInt(c2, 16)).toString(16);

  while (hexStr.length < 6) { hexStr = '0' + hexStr; } // Zero pad.

  return hexStr;

}



addHexColor('aaaaaa', '010101'); // => 'ababab'

addHexColor('010101', '010101'); // => '020202'

As mentioned by a commenter, the above solution is chock full of problems, so below is a function that does proper input validation and adds color channels separately while checking for overflow.

function addHexColor2(c1, c2) {

  const octetsRegex = /^([0-9a-f]{2})([0-9a-f]{2})([0-9a-f]{2})$/i

  const m1 = c1.match(octetsRegex)

  const m2 = c2.match(octetsRegex)

  if (!m1 || !m2) {

    throw new Error(`invalid hex color triplet(s): ${c1} / ${c2}`)

  }

  return [1, 2, 3].map(i => {

    const sum = parseInt(m1[i], 16) + parseInt(m2[i], 16)

    if (sum > 0xff) {

      throw new Error(`octet ${i} overflow: ${m1[i]}+${m2[i]}=${sum.toString(16)}`)

    }

    return sum.toString(16).padStart(2, '0')

  }).join('')

}



addHexColor2('aaaaaa', 'bogus!') // => Error: invalid hex color triplet(s): aaaaaa / bogus!

addHexColor2('aaaaaa', '606060') // => Error: octet 1 overflow: aa+60=10a