Sass - a way to invert Lighten/Darken function by variable? - sass

Question

I'm trying to think of a simple way to invert the darken/lighten color function using a variable. Something like so...
$invert_switch: off;
And...
#if $invert_switch == "on" {
darken == lighten
lighten == darken
} #else {
darken == darken
lighten == lighten
}
I know that's not proper code, I'm just trying to explain it very simply.
If this was used...
background: darken($bg, 10%);
color: lighten($txt, 50%);
How could I flip that, so it would be...
background: lighten($bg, 10%);
color: darken($txt, 50%);
Or maybe there is an easier way?

Answer 1

I try to change the name of the functions darken and lighten but I can't. The problem I have found is that SASS haven't something like #break or #continue to exit for the function and do nothing, my idea was the following code but when the custom fuction coincides with the SASS function goes to a infinite loop:
$invert_switch: on;
#function darken($color, $amount){
#if $invert_switch == on {
#return lighten($color, $amount);
}
#else {
#return darken($color, $amount);
}
}
#function lighten($color, $amount){
#if $invert_switch == on {
#return darken($color, $amount);
}
#else {
#return lighten($color, $amount);
}
}
h1{
color: darken(red, 10%);
}
p{
color: lighten(red, 10%);
}
I change the name of the functions adding a underscore to work:
$invert_switch: on;
#function _darken($color, $amount){
#if $invert_switch == on {
#return lighten($color, $amount);
}
#else {
#return darken($color, $amount);
}
}
#function _lighten($color, $amount){
#if $invert_switch == on {
#return darken($color, $amount);
}
#else {
#return lighten($color, $amount);
}
}
h1{
color: _darken(red, 10%);
}
p{
color: _lighten(red, 10%);
}
Maybe someone can improve this code using the proper names of functions