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? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

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.