I have setup several functions in my .bashrc file. I would like to just display the actual code of the function and not execute it, to quickly refer to something.
Is there any way, we could see the function definition?
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4 Answers
The declare builtin's -f option does that:
bash-4.2$ declare -f apropos1 apropos1 () { apropos "$@" | grep ' (1.*) ' } I use type for that purpose, it is shorter to type ;)
bash-4.2$ type apropos1 apropos1 is a function apropos1 () { apropos "$@" | grep ' (1.*) ' } 
- In Ubuntu 18.04,
typeoutputs:<function> is a shell function from <location>, and not the content.Timo– Timo2020-04-07 04:54:37 +00:00Commented Apr 7, 2020 at 4:54 - 2@Timo, are you sure that is Bash? Sounds like Zsh behavior.manatwork– manatwork2020-04-07 09:03:30 +00:00Commented Apr 7, 2020 at 9:03
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You can use the type command to do this.
type yourfunc will print the function to STDOUT. As man type says,
The type utility shall indicate how each argument would be interpreted if used as a command name.
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man type? Shell builtins usually have no man page on my system.manatwork– manatwork2012-12-27 08:41:26 +00:00Commented Dec 27, 2012 at 8:41 - I have updated my answer with a link. The system I use is Arch Linux.jasonwryan– jasonwryan2012-12-27 08:55:37 +00:00Commented Dec 27, 2012 at 8:55
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- No problem: as it is one of the POSIX man pages, I thought it would be widely distributed...jasonwryan– jasonwryan2012-12-27 09:10:57 +00:00Commented Dec 27, 2012 at 9:10
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help typeshows the Bash-specific information.l0b0– l0b02012-12-27 12:25:18 +00:00Commented Dec 27, 2012 at 12:25
for builtin commands' info use:
help [-s|-d] COMMAND1 COMMAND2 .... for example:
help help alias For info about all of them type, for example:
help -s '' 
type works if you declared your function in the shell but which works even if you sourced the function from another file.
- That's not true.
typeworks fine no matter where you've sourced the function from. Tryprintf 'foo(){ echo "hi"; }\n' > file; . file; type foo. Actually,whichdoesn't work with functions at all, it only finds commands in your PATH. See Why not use "which"? What to use then?2022-11-02 09:54:20 +00:00Commented Nov 2, 2022 at 9:54 - @terdon, depends on the implementation of
which, GNUwhichcan be configured to dump function definitions (as you'll see at the Q&A you're linking to)Stéphane Chazelas– Stéphane Chazelas2022-11-02 10:27:15 +00:00Commented Nov 2, 2022 at 10:27 - @StéphaneChazelas even there, you would need something like
declare -f | which --read-functions --tty-only $functionNameand it still makes no difference whether the function was defined in the current shell or sourced from a file.2022-11-02 10:40:35 +00:00Commented Nov 2, 2022 at 10:40