Bash functions ignore errors when called in a boolean expression?

Question

The following bash script prints "This should never run!":

#!/bin/bash
set -e

func() {
    echo "Bailing out..."
    false # Should bail out because of set -e

    echo "This should never run!"
}

func || true

What's going on here?

A simple conclusion to make is that set -e is not propagated inside functions. However, this is not the case. Changing the last line (func || true) to just func causes the function to stop at the false statement as expected.

Another hypothesis is that bash will start executing the function, and on the first error (false) evaluate the rest of the line it was called on (func || true), and if that retuns 0 (true), then it continues executing the function as if nothing happened. However, replacing true with echo "Oops!" disproves that, since no "Oops!" is printed.

Is set -e somehow ignored when executing functions as part of a boolean expression?

Answer 1

From man bash regarding set -e:

Exit immediately if a pipeline (which may consist of a single simple command), a list, or a compound command (see SHELL GRAMMAR above), exits with a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test following the if or elif reserved words, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command's return value is being inverted with !. If a compound command other than a subshell returns a non-zero status because a command failed while -e was being ignored, the shell does not exit. A trap on ERR, if set, is executed before the shell exits. This option applies to the shell environment and each subshell environment separately (see COMMAND EXECUTION ENVIRONMENT above), and may cause subshells to exit before executing all the commands in the subshell.