Is trap inherited by a subshell?

Question

I tried a following script:

#!/bin/bash
trap 'echo "touching a file" && touch $FILE' EXIT

foo1(){
        echo "foo1"
}
foo(){
        echo "foo"
        export FILE=${FILE:-/tmp/file1}
}
(foo1)
foo

The output for the above script was:

[root@usr1 my_tests]# ./test.sh
foo1
foo
touching a file

However I was expecting trap to be called on exit from foo1 as well, which is called in a subshell.

• Is this expected?
• Is trap inherited by a subshell?
• If yes, then in what case is trap inherited by a subshell?

Answer 1

Trap handlers are never inherited by subshells. This is specified by POSIX:

When a subshell is entered, traps that are not being ignored are set to the default actions.

Note that ignored signals (trap '' SIGFOO) remain ignored in the subshell (and in external programs launched by the shell, too).

Answer 2

trap is not propogated to subshells but some ways allow the subshell to report the traps of the parent shell and others don't. I did some tests on macos with bash.

GNU bash, version 4.4.12(1)-release (x86_64-apple-darwin16.3.0):

trap 'echo hello' EXIT
trap # trap -- 'echo hello' EXIT
echo "$(trap)" # trap -- 'echo hello' EXIT
trap | cat # trap -- 'echo hello' EXIT
(trap) | cat # trap -- 'echo hello' EXIT
cat < <(trap) # empty
cat <<< "$(trap)" # empty
bash -c 'trap' # empty
trap & # trap -- 'echo hello' EXIT

GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin16):

trap 'echo hello' EXIT
trap # trap -- 'echo hello' EXIT
echo "$(trap)" # trap -- 'echo hello' EXIT
trap > >(cat) # trap -- 'echo hello' EXIT
trap | cat # empty
(trap) | cat # empty
cat < <(trap) # empty
cat <<< "$(trap)" # empty
bash -c 'trap' # empty
trap & # empty

This is good to know that trap_output="$(trap)" will work to capture trap output. I can't think of any other way to do it if that didn't work besides doing trap >trap_output_file to output it to a file (fifo won't work in bash 3.2.57) and then reading it back in with trap_output="$(<trap_output_file)"

fifo won't work in bash 3.2.57 because trap & is empty for bash 3.2.57 but not bash 4.4.12

GNU bash, version 4.4.12(1)-release (x86_64-apple-darwin16.3.0):

mkfifo /tmp/fifo; trap >/tmp/fifo & trap_output=$(</tmp/fifo); rm -f /tmp/fifo; echo "$trap_output"
# trap -- 'echo hello' EXIT

mkfifo /tmp/fifo; trap_output=$(</tmp/fifo) & trap >/tmp/fifo; rm -f /tmp/fifo; echo "$trap_output"
# empty because trap_output=$(</tmp/fifo) sets the variable in a subshell

GNU bash, version 3.2.57(1)-release (x86_64-apple-darwin16):

mkfifo /tmp/fifo; trap >/tmp/fifo & trap_output=$(</tmp/fifo); rm -f /tmp/fifo; echo "$trap_output"
# empty because trap >/tmp/fifo & is empty since it uses trap &

mkfifo /tmp/fifo; trap_output=$(</tmp/fifo) & trap >/tmp/fifo; rm -f /tmp/fifo; echo "$trap_output"
# empty because trap_output=$(</tmp/fifo) sets the variable in a subshell

Answer 3

trap definitions are not propagated to sub-shells.

Verify by:

trap "echo bla" 1 2 3

(trap)

Answer 4

Although calling trap in a subshell shows the parent's traps, they are not executed in the subshell. You can set a new trap in the subshell.

To test this, create a script with the following code, run it and ctrl+c:

#!/usr/bin/bash
function sub { echo sub_before: `trap`; trap 'echo kill_sub; exit' SIGINT; echo sub_after: `trap`; while :; do sleep 1; done; }
function parent { echo parent_before: `trap`; trap 'pkill -2 -P $$; echo kill_parent; exit' SIGINT; echo parent_after: `trap`; sub & wait; }
parent

You'll see both traps being executed.

(bash 5.0.17)