Bash: Why does parent script not terminate on SIGINT when child script traps SIGINT?

New answer:

This question is far more interesting than I originally suspected. The answer is essentially given here:

What happens to a SIGINT (^C) when sent to a perl script containing children?

Here's the relevant tidbit. I realize you're not using Perl, but I assume Bash is using C's convention.

Perl’s builtin system function works just like the C system(3) function from the standard C library as far as signals are concerned. If you are using Perl’s version of system() or pipe open or backticks, then the parent — the one calling system rather than the one called by it — will IGNORE any SIGINT and SIGQUIT while the children are running.

This explanation is the best I've seen about the various choices that can be made. It also says that Bash does the WCE approach. That is, when a parent process receives SIGINT, it waits until its child process returns. If that process handled exited from a SIGINT, it also exits with SIGINT. If the child exited any other way it ignores SIGINT.

There is also a way that the calling shell can tell whether the called program exited on SIGINT and if it ignored SIGINT (or used it for other purposes). As in the WUE way, the shell waits for the child to complete. It figures whether the program was ended on SIGINT and if so, it discontinue the script. If the program did any other exit, the script will be continued. I will call the way of doing things the "WCE" (for "wait and cooperative exit") for the rest of this document.

I can't find a reference to this in the Bash man page, but I'll keep looking in the info docs. But I'm 99% confident this is the correct answer.

Old answer:

A nonzero exit status from a command in a Bash script does not terminate the program. If you do an echo $? after ./script2.sh it will show 130. You can terminate the script by using set -e as phs suggests.

$ help set
...
-e  Exit immediately if a command exits with a non-zero status.

The second part of @seanmcl's updated answer is correct and the link to http://www.cons.org/cracauer/sigint.html is a really good one to read through carefully.

From that link, "You cannot 'fake' the proper exit status by an exit(3) with a special numeric value, even if you look up the numeric value for your system". In fact, that's what is being attempted in @Hermann Speiche's script2.sh.

One answer is to modify function handler in script2.sh as follows:

function handler {
  # ... do stuff ...
  trap INT
  kill -2 $$
}

This effectively removes the signal handler and "rethrows" the SIGINT, causing the bash process to exit with the appropriate flags such that its parent bash process then correctly handles the SIGINT that was originally sent to it. This way, using set -e or any other hack is not actually required.

It's also worth noting that if you have an executable that behaves incorrectly when sent a SIGINT (it doesn't conform to "How to be a proper program" in the above link, e.g. it exits with a normal return-code), one way of working around this is to wrap the call to that process with a script like the following:

#!/bin/bash

function handler {
  trap INT
  kill -2 $$
}

trap handler INT
badprocess "$@"