Can I “export” functions in bash?

Question

source some_file

some_file:

doit ()
{
  echo doit $1
}
export TEST=true

If I source some_file the function "doit" and the variable TEST are available on the command line. But running this script:

script.sh:

#/bin/sh
echo $TEST
doit test2

Will return the value of TEST, but will generate an error about the unknown function "doit".

Can I "export" the function, too, or do I have to source some_file in script.sh to use the function there?

Answer 1

You cannot export functions, not in the way that you are describing. The shell will only load the ~/.bashrc file on the start of an interactive shell (search for "Invocation" in the bash manpage).

What you can do is create "library" which is loaded when you start the program:

source "$HOME/lib/somefile"

And place your non-interactive functions and settings there.

Answer 2

In bash you can export function definitions to sub-shell with

export -f function_name

For example you can try this simple example:

./script1:

    #!/bin/bash

    myfun() {
        echo "Hello!"
    }

    export -f myfun
    ./script2

./script2:

    #!/bin/bash

    myfun

Then if you call ./script1 you will see the output Hello!.

Answer 3

Functions are not exported to subprocesses. This is why there are files named .kshrc or .bashrc: To define functions that shoiuld be available in subshells also.

If running a script, the .*shrc scripts are normally not sourced. You would have to code that explicitly, like in . ~/.kshrc.