Is there a way to write a bash function which aborts the whole execution, no matter how it is called?

Question

I was using \"exit 1\" statement in my bash functions to terminate the whole script and it worked fine:

function func()
{
   echo \"Goodbye\"
   exit 1
}
ech         


        

Answer 1

What you could do, is register the top level shell for the TERM signal to exit, and then send a TERM to the top level shell:

#!/bin/bash
trap "exit 1" TERM
export TOP_PID=$$

function func()
{
   echo "Goodbye"
   kill -s TERM $TOP_PID
}

echo "Function call will abort"
echo $(func)
echo "This will never be printed"

So, your function sends a TERM signal back to the top level shell, which is caught and handled using the provided command, in this case, "exit 1".

Answer 2

You can use set -e which exits if a command exits with a non-zero status:

set -e 
func
set +e

Or grab the return value:

(func) || exit $?

Answer 3

I guess better is

#!/bin/bash
set -e
trap "exit 1" ERR

myfunc() {
     set -x # OPTIONAL TO SHOW ERROR
     echo "Exit with failure"
     set +x # OPTIONAL
     exit 1
}
echo "BEFORE..."
myvar="$(myfunc)"
echo "AFTER..But not shown"

Answer 4

A child process can't force the parent process to close implicitly. You need to use some kind of signaling mechanism. Options might include a special return value, or perhaps sending some signal with kill, something like

function child() {
    local parent_pid="$1"
    local other="$2"
    ...
    if [[ $failed ]]; then
        kill -QUIT "$parent_pid"
    fi
}

Answer 5

But is there a way to write a function which aborts the whole execution, no matter how it is called?

No.

I just need to get the real return value (echoed by the function).

You can

res=$(func)
echo $?