trap in exported function silently ignored

Question

This Bash script behaves as expected.

test_this.sh

function run_this() {
    trap "echo TRAPPED" EXIT
    false
    echo $?
}
run_this

It prints

1
TRAPPED

However, when I try to export this function, it fails to trap.

test_this2.sh

function run_this() {
    trap "echo TRAPPED" EXIT
    false
    echo $?
}
export -f run_this

Source this at the command line and run it:

> source test_this2.sh
> run_this

Results in

Where did the trap go?

David C. Rankin · Accepted Answer · 2014-11-26T02:21:33.587

The trap is ignored when you export the function because when you exit from your login shell (where the function is exported to), there is no longer a shell to print trapped in. (i.e. there is never an exit otherwise you would no longer have a shell.) When you source test_this2.sh, you execute it in your login shell. When the function completes, it returns to your login shell -- there is no exit. When you run test_this.sh, it executes in a subshell, when the subshell exits, you get trapped printed. If you really want to see what happens when you exit your login shell, try typing exit and see what happens.

score 1 · Answer 2 · answered Apr 11 '16 at 09:36

If you want to 'trap' leaving the function use the RETURN signal. Use the ERR signal to 'trap' errors like your 'false' call.

test_this.sh

function run_this() {
    trap "echo LEAVING" RETURN
    trap "echo ERROR" ERR
    false
    echo $?
}
export -f run_this

Load and export the function

source this_this.sh

test it

run_this

returns

ERROR
1
LEAVING

start subshell

bash

test again

run_this

returns

ERROR
1
LEAVING

So the export works fine.

PS: I'm using bash 4.3.42

trap in exported function silently ignored

2 Answers2