When Comment Out 'exit', Code Perform Well

Question

I am following an instruction and confuse with the usage of exit

    FILE=~/.bashrc
    if [ -e "$FILE" ] ; then
        if [ -f "$FILE" ]; then
            echo "$FILE is a regular file."
        fi
        if [ -d "$FILE" ]; then
            echo "$FILE is a directory."
        fi
        if [ -r "$FILE" ] ; then
            echo "$FILE is readable."
        fi
        if [ -w "$FILE" ] ; then
            echo "$FILE is writabe."
        fi
        if [ -x "$FILE" ]; then
            echo '$FILE is executable/searchable.'
        fi
        else
            echo '$FILE does not exist'
            exit 1
        fi
    exit

Run and come by

    $ bash test_file.sh
    /Users/me/.bashrc is a regular file.
    /Users/me/.bashrc is readable.
    /Users/me/.bashrc is writabe.

If comment out commands of exit, the output stay unchanged.

What's its function? Could I leave out the exits when familiar with the language.

Answer 1

The exit command not only exits a script, but also sets an exit code, which by convention is zero for a successful exit, and some other integer for an error, so in your script, its purpose is to indicate to the caller (either another script / program, or a user) that the program exited with an error. In bash and similar shells, one can view or interrogate the exit code by examining shell variable $?.

Also, BTW, you have an indentation problem. The else clause should be out-dented to the same level as the initial if statement...