Parent trap visible but not run by subshell

Question

Tested for Bash 5.0.2

According to the GNU Bash Reference Manual,

Bash performs the expansion [of a command substitution] by executing [the] command in a subshell environment

According to The Open Group Base Specifications Issue 6:

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

---

So when running the following script:

function a {
   trap -p EXIT
}

trap "echo 'parent'" EXIT

echo "$(a)"
(a)

trap - EXIT

echo 'exiting'

... i would expect an output of:

exiting

... but instead I get:

trap -- 'echo '\''parent'\''' EXIT
trap -- 'echo '\''parent'\''' EXIT
exiting

... meaning that the function a - eventhough it is being run in a subshell - is seeing the the parent shell's trap commands (via trap -p) but not executing them.

---

What is going on here?

Answer 1

You'll notice that the traps trigger exactly according to spec. It's just the output from trap that's unexpected.

This is a feature in Bash 4.2 (release notes):

b.  Subshells begun to execute command substitutions or run shell functions or
    builtins in subshells do not reset trap strings until a new trap is
    specified.  This allows $(trap) to display the caller's traps and the
    trap strings to persist until a new trap is set.

Normally, people would take this for granted. Consider this totally unsurprising Bash exchange:

bash$ trap
trap -- 'foo' EXIT
trap -- 'bar' SIGINT

bash$ trap | grep EXIT
trap -- 'foo' EXIT

Now look at the result in other shells like Dash, Ksh or Zsh:

dash$ trap
trap -- 'foo' EXIT
trap -- 'bar' INT

dash$ trap | grep EXIT
(no output)

This is perhaps more correct, but I doubt many people would expect it.