How to trap ERR when using 'set -e' in Bash

chepner's answer is the best solution: If you want to combine set -e (same as: set -o errexit) with an ERR trap, also use set -o errtrace (same as: set -E).

In short: use set -eE in lieu of just set -e:

#!/bin/bash

set -eE  # same as: `set -o errexit -o errtrace`
trap 'echo BOO!' ERR 

function func(){
  ls /root/
}

# Thanks to -E / -o errtrace, this still triggers the trap, 
# even though the failure occurs *inside the function*.
func 

^{A more sophisticated example trap example that prints the message in red and also prints the exit code:
trap 'printf "\e[31m%s: %s\e[m\n" "BOO!" $?' ERR}

---

man bash says about set -o errtrace / set -E:

If set, any trap on ERR is inherited by shell functions, command substitutions, and commands executed in a subshell environment. The ERR trap is normally not inherited in such cases.

What I believe is happening:

• Without -e: The ls command fails inside your function, and, due to being the last command in the function, the function reports ls's nonzero exit code to the caller, your top-level script scope. In that scope, the ERR trap is in effect, and it is invoked (but note that execution will continue, unless you explicitly call exit from the trap).

• With -e (but without -E): The ls command fails inside your function, and because set -e is in effect, Bash instantly exits, directly from the function scope - and since there is no ERR trap in effect there (because it wasn't inherited from the parent scope), your trap is not called.

While the man page is not incorrect, I agree that this behavior is not exactly obvious - you have to infer it.

You need to use set -o errtrace for the function to inherit the trap.