SASS mixing and passing a variable string that's a function name - sass

Question

I'm expanding out a SASS mixin for generating background colours.
Here's my mixin:
#mixin color-columns ($count, $baseName, $startcolor) {
$loop_color : $startcolor;
#for $i from 0 through $count {
$loop_color : darken($loop_color, 8%);
.#{$baseName}-#{$i} {
background-color : $loop_color;
}
}
}
And i use it like this:
#include color-columns(5,'col', $blue);
What I want to do is pass in either 'darken' or 'lighten' as a variable so i can control with SASS colour function. Something like this:
#mixin color-columns ($count, $baseName, $startcolor, $change: 'darken') {
$loop_color : $startcolor;
#for $i from 0 through $count {
$loop_color : $change($loop_color, 8%);
.#{$baseName}-#{$i} {
background-color : $loop_color;
}
}
}
When i try that, my CSS for the first column is like this:
background-color: darken darken #005387, 8%, 8%;
I'm missing the connection on passing a variable that swaps the function.

Answer 1

Currently there is no support for defining dynamic function names. It has been requested several years ago and there is even an open pull request for this functionality but it is not yet merged.
However, it is possible to dynamically call a function with the built-in call($function, $args...) function. It allows you to avoid if-else cascades and can be used in your example like this:
$loop_color : call($change, $loop_color, 8%);

Answer 2

I ended up adding an #if to get this to work:
#mixin color-columns ($count, $startcolor, $start : darken) {
$loop_color : $startcolor;
#if $start == 'lighten' {
#for $i from 0 through $count {
$loop_color : lighten($loop_color, 5%);
.conBG-#{$i} {
background-color : $loop_color;
}
}
}
#else {
#for $i from 0 through $count {
$loop_color : darken($loop_color, 5%);
.conBG-#{$i} {
background-color : $loop_color;
}
}
}
}