bash: Why can't I set a trap for SIGINT in a background shell?

Question

Here's a simple program that registers two trap handlers and then displays them with trap -p. Then it does the same thing, but in a child background process.

Why does the background process ignore the SIGINT trap?

#!/bin/bash

echo "Traps on startup:"
trap -p
echo ""

trap 'echo "Received INT"' INT
trap 'echo "Received TERM"' TERM

echo "Traps set on parent:"
trap -p
echo ""

(
    echo "Child traps on startup:"
    trap -p
    echo ""

    trap 'echo "Child received INT"' INT
    trap 'echo "Child received TERM"' TERM

    echo "Traps set on child:"
    trap -p
    echo ""
) &

child_pid=$!
wait $child_pid

Output:

$ ./show-traps.sh
Traps on startup:

Traps set on parent:
trap -- 'echo "Received INT"' SIGINT
trap -- 'echo "Received TERM"' SIGTERM

Child traps on startup:

Traps set on child:
trap -- 'echo "Child received TERM"' SIGTERM

Answer 1

SIGINT and SIGQUIT are ignored in backgrounded processes (unless they're backgrounded with set -m on). It's a (weird) POSIX requirement (see http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html or my SO question Why do shells ignore SIGINT and SIGQUIT in backgrounded processes? for more details).

Additionally, POSIX requires that:

When a subshell is entered, traps that are not being ignored shall be set to the default actions, except in the case of a command substitution containing only a single trap command ..

However, even if you set the INT handler in the subshell again after it was reset, the susbshell won't be able to receive it because it's ignored (you can try it or you can inspect the signal ignore mask using ps, for example).