Display the function body in Bash

Question

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?

score 29 · Accepted Answer · answered Dec 27 '12 at 08:38

29

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.*) '
}

answered Dec 27 '12 at 08:38

manatwork

31,277

In Ubuntu 18.04, type outputs: <function> is a shell function from <location>, and not the content. – Timo Apr 07 '20 at 04:54
1

@Timo, are you sure that is Bash? Sounds like Zsh behavior. – manatwork Apr 07 '20 at 09:03
thanks for the hint, here is the solution. – Timo Apr 08 '20 at 05:07

jasonwryan · Answer 2 · 2012-12-27T08:53:34.900

7

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.

edited Dec 27 '12 at 08:53

answered Dec 27 '12 at 08:38

jasonwryan

73,126

3

man type? Shell builtins usually have no man page on my system. – manatwork Dec 27 '12 at 08:41
I have updated my answer with a link. The system I use is Arch Linux. – jasonwryan Dec 27 '12 at 08:55
1

Thank you. Seems the man-pages package contains much more on Arch. – manatwork Dec 27 '12 at 09:05
No problem: as it is one of the POSIX man pages, I thought it would be widely distributed... – jasonwryan Dec 27 '12 at 09:10
5

help type shows the Bash-specific information. – l0b0 Dec 27 '12 at 12:25

score 0 · Answer 3 · answered Feb 21 '14 at 05:16

0

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 ''

answered Feb 21 '14 at 05:16

Robert

1

score -1 · Answer 4 · edited Nov 02 '22 at 09:52

-1

type works if you declared your function in the shell but which works even if you sourced the function from another file.

edited Nov 02 '22 at 09:52

Fco Javier Balón

1,173

answered Nov 01 '22 at 20:59

Tom

1

That's not true. type works fine no matter where you've sourced the function from. Try printf 'foo(){ echo "hi"; }\n' > file; . file; type foo. Actually, which doesn't work with functions at all, it only finds commands in your PATH. See Why not use "which"? What to use then? – terdon Nov 02 '22 at 09:54
@terdon, depends on the implementation of which, GNU which can be configured to dump function definitions (as you'll see at the Q&A you're linking to) – Stéphane Chazelas Nov 02 '22 at 10:27
@StéphaneChazelas even there, you would need something like declare -f | which --read-functions --tty-only $functionName and it still makes no difference whether the function was defined in the current shell or sourced from a file. – terdon Nov 02 '22 at 10:40

Display the function body in Bash

4 Answers4