Determine if a function exists in bash


Currently I'm doing some unit tests which are executed from bash. Unit tests are initialized, executed and cleaned up in a bash script. This script usualy contains an init(), execute() and cleanup() functions. But they are not mandatory. I'd like to test if they are or are not defined.

I did this previously by greping and seding the source, but it seemed wrong. Is there a more elegant way to do this?

Edit: The following sniplet works like a charm:

    LC_ALL=C type $1 | grep -q 'shell function'
3/6/2019 8:08:43 PM

Accepted Answer

I think you're looking for the 'type' command. It'll tell you whether something is a function, built-in function, external command, or just not defined. Example:

$ LC_ALL=C type foo
bash: type: foo: not found

$ LC_ALL=C type ls
ls is aliased to `ls --color=auto'

$ which type

$ LC_ALL=C type type
type is a shell builtin

$ LC_ALL=C type -t rvm

$ if [ -n "$(LC_ALL=C type -t rvm)" ] && [ "$(LC_ALL=C type -t rvm)" = function ]; then echo rvm is a function; else echo rvm is NOT a function; fi
rvm is a function
2/16/2019 3:57:41 AM

$ g() { return; }
$ declare -f g > /dev/null; echo $?
$ declare -f j > /dev/null; echo $?

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