Selective random colors in JavaScript

Question

Given the following code to generate random colors (credit):

'#'+((1<<24)*(Math.random()+1)|0).toString(16).substr(1)

What conditions can be added to this code, in order to make it produce only bright colors but darker than pure white #ffffff.

The elements which are assigned these colors, need to be displayed on a dark background (#111).

What is your brightness threshold? You would basically try to determine if each pair of hex numbers are < ff and > some brightness threshold or 00. — Scottux
– Scottux, Commented Dec 8, 2014 at 20:54

Cara McCormack · Accepted Answer · 2014-12-08 20:57:46Z

2

Rather than messing with the bitwise left-shift, it's probably easier to generate each digit separately, like so:

var color = '#';
for (var i = 0; i < 3; i+=1) {
  var first = Math.floor((Math.random()*8+8)).toString(16);
  var second = Math.floor((Math.random()*16)).toString(16);

  color = color + first + second;
}

If you want to adjust exactly how light/dark the colors are, change the Math.random()*8+8 to adjust the first digit in each of the R, G, B values. As it stands, this will generate anything from #808080 to #ffffff. If you wanted from #a0a0a0 to #ffffff, change it *10+6 - it can be anything in the form *x+y so long as x and y add up to 16.

answered Dec 8, 2014 at 20:57

Cara McCormack

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1 Comment

Annie Over a year ago

Thank you! Not one-liner, but this code is more clear and readable. :)

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