What does set -e mean in a bash script?

Question

I'm studying the content of this **preinst** file that the script executes before that package is unpacked from its Debian archive (.deb) file. The script has the following code: #!/bin/bash set -e # Automatically added by dh_installinit if [ "$1" = install ]; then if [ -d /usr/share/MyApplicationName ]; then echo "MyApplicationName is just installed" return 1 fi rm -Rf $HOME/.config/nautilus-actions/nautilus-actions.conf rm -Rf $HOME/.local/share/file-manager/actions/* fi # End automatically added section My first query is about the line: set -e I think that the rest of the script is pretty simple: It checks whether the Debian/Ubuntu package manager is executing an install operation. If it is, it checks whether my application has just been installed on the system. If it has, the script prints the message **"MyApplicationName is just installed"** and ends (`return 1` mean that ends with an “error”, doesn’t it?). If the user is asking the Debian/Ubuntu package system to install my package, the script also deletes two directories. Is this right or am I missing something?

Answer 1

set -e means exit immediately if anything you're running returns a non-zero return code (meaning that something didn't run successfully).

The section if [ "$1" = install ] is simple as well. In shell scripting, you can reference parameters passed to the script with a dollar sign "$" and then the number of the param. So in the case above is saying "if the value of the first parameter passed to the script equals install, then verify if the directory /usr/share/MyApplicationName exists.

So if you were to run: "./script.sh hello there", $1 would have the value "hello" while $2 would be "there".