Answer by entpnerd for What does 'set -e' mean in a Bash script?

I found this post while trying to figure out what the exit status was for a script that was aborted due to a set -e. The answer didn't appear obvious to me; hence this answer. Basically, set -e aborts the execution of a command (e.g. a shell script) and returns the exit status code of the command that failed (i.e. the inner script, not the outer script).

For example, suppose I have the shell script outer-test.sh:

#!/bin/shset -e./inner-test.shexit 62;

The code for inner-test.sh is:

#!/bin/shexit 26;

When I run outer-script.sh from the command line, my outer script terminates with the exit code of the inner script:

$ ./outer-test.sh$ echo $?26