In the Z shell there's a handy command that returns a list of all available functions. The command is, conveniently, called functions. I cannot find a similar alternative in Bash. I threw together a quick & dirty (and wholly unacceptable) function to approximately do the same thing, but it has at least one glaring problem: since it relies on parsing files you must either list all the files to look in (which may become stale) or give an expression (which is guaranteed to give files you don't want to look in, such as .bash_history).
Here's the function, since I know someone will ask for it if I don't post it, but I'm pretty sure it's a dead end, or at least the wrong approach.
functions() {
grep "^function " "$HOME/."{bashrc,bash_profile,aliases,functions,projects,variables} | sort | sed -e 's/{//' | uniq
}
I could improve on this wrong-headed approach by parsing .bash_profile and getting a list of all sourced files and then parsing them for functions, but by the time you add the following complications into the mix, it's really not worth it:
- You can source files with
.orsource. - I also happen to use a function to source files, which checks for the file's existence first.
- You could easily source after
&∨: it's not necessarily the first or only thing on a line. - You have to account for the fact that functions don't necessarily have the keyword
functionbefore them. - You can omit the () after the function name.
- There are probably other complicating factors I haven't thought of.
- Fundamentally this is wrong because it is parsing files rather than reporting what is loaded in memory.
Is there any reasonable way to do this—get a list of all functions loaded in memory—in Bash? It seems like an enormous omission, if not.
(And for those looking for duplicate questions, this one is very different, as it's asking for a way to list only those functions that come from a specific file.)
