What does set -e mean in a bash script?

From help set :

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

But it's considered bad practice by some (bash FAQ and irc freenode #bash FAQ authors). It's recommended to use:

trap 'do_something' ERR

to run do_something function when errors occur.

See http://mywiki.wooledge.org/BashFAQ/105

set -e stops the execution of a script if a command or pipeline has an error - which is the opposite of the default shell behaviour, which is to ignore errors in scripts. Type help set in a terminal to see the documentation for this built-in command.

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