How to output rgb(n,n,n) ?

I am struggling to get this to work as expected:

$color-black = rgb(0,0,0) // #000000
$color-white = rgb(255,255,255) // #FFFFFF
$color-red = rgb(201,37,49); // #c92531


$colors = primary-red $color-red, primary-white $color-white, primary-black $color-black;

generatestuff($c)
  for $pair in $c
   .color-{$pair[0]}
      color: rgb($pair[1])
generatestuff($colors)

The output is:

.color-primary-red {
  color: #c92531;
}
.color-primary-white {
  color: #fff;
}
.color-primary-black {
  color: #000;
}

What I expect/wants/desire…

.color-primary-red {
  color: rgb(201,37,49);
}
.color-primary-white {
  color: rgb(255,255,255);
}
.color-primary-black {
  color: rgb(0,0,0);
}

How can I achieve this if this is not intended behaviour?

[Panya]

This is intended behaviour. The only way you can achieve this is by saving r, g, b values as lists and use unquote('rgb(' + join(',', $color) + ')'):

$color = 0, 0, 0

body
  color: unquote('rgb(' + join(',', $color) + ')')

[phun-ky]

@Panya @srsgores Sorry for bringing up an old issue, encountered the use case again today.

I was trying to interpolate color variables:

From:

a
  border-bottom-color rgba(37, 116, 168, 0.9)
  background-image linear-gradient(180deg,rgba(37, 116, 168, .4) 0,transparent 1px,transparent)

To:

$color-a = rgb(37, 116, 168)
$size-b = 1px
a
  border-bottom-color rgba($color-a, 0.9)
  background-image "linear-gradient(180deg,rgba(%s, .4) 0,transparent %s,transparent)" % ($color-a $size-b)

And since the border-bottom-color color works, why wouldn’t it work in string interpolation. But it struck me, the interpolation is sent straight trough the parser, right?

Because this is the desired outcome:

a{
  border-bottom-color: rgba(37, 116, 168, 0.9);
  background-image: linear-gradient(180deg,rgba(37, 116, 168, .4) 0,transparent 1px,transparent);
}

But we get this:

a{
  border-bottom-color: rgba(37, 116, 168, 0.9);
  background-image: linear-gradient(180deg,rgba(#2574a8, .4) 0,transparent 1px,transparent);
}

One way is to do this:

$color-a = rgb(37, 116, 168)
$size-b = 1px
a
  border-bottom-color rgba($color-a, 0.9)
  background-image "linear-gradient(180deg,%s 0,transparent %s,transparent)" % (rgba($color-a, .4) $size-b)

But it’s not very readable.

And this does not either work:

$color-a = rgb(37, 116, 168)
a
  color rgba($color-a, 0.9) // results in rgba(37, 116, 168, 0.9)
  color rgba($color-a, 1) // results in #2574a8

Which makes things a bit inconsistent when searching and debugging the code.

Dunno if this is a rant or a real issue, but needed to jot down my thoughts about this…