export function with different name

Question

I have a script that exports a function for child scripts to use and would like to change what function is exported based on a conditional. I know I could just declare the two different functions within the conditional but would prefer to avoid that if possible.

So an example would be:

#!/bin/bash

foo () { echo foo; }
bar () { echo bar; }

if [[ $var == foo ]]; then
    #in this case my_func() should execute foo()
    export -f my_func
elif [[ $var == bar ]]; then
    #in this case my_func() should execute bar()
    export -f my_func
fi

I guess I could also just make a simple inception function like my_func() { foo "$@"; } inside the conditional before the export too but am asking if there is a better way to do this.

Just trying things, not that I know what I'm doing... foo () { echo foo; }; eval "$(echo "my_func()"; declare -f foo | tail -n +2)"; my_func → foo (export -f closes my terminal). — schrodingerscatcuriosity
– schrodingerscatcuriosity, Commented Feb 3, 2022 at 16:37

Kusalananda · Accepted Answer · 2022-02-03 18:55:56Z

This may be what you were already proposing, but I don't think there's a significantly better way. The shell doesn't provide a safe introspective way of including the actual declaration of the pre-declared foo or bar functions in the declaration of my_func, which I suppose would be an alternative. And if you're not willing to move the functionality of foo or bar into my_func (i.e., without ever creating foo or bar as separate functions), then what I show here may be best:

#!/bin/bash

foo () { echo foo; }
bar () { echo bar; }

case $var in
    foo) export -f foo; my_func () { foo "$@"; } ;;
    bar) export -f bar; my_func () { bar "$@"; } ;;
esac

export -f my_func

That is, depending on the value of var, export the appropriate function(s) that your my_func is using and declare your function. Then export it.

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