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?

Answer 1

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.*) '
}

Answer 2

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.

Answer 3

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 ''