Evaluating command inside bash script leads to increment in $SHLVL

Question

I have the following bash script (myscript.sh):

#!/bin/bash
$(echo $0)

Now: before sourcing myscript.sh, $SHLVL is 2. After sourcing it, is 3. Why? Shouldn't the command evaluation exit after echo?

Thank you!

Answer 1

When you source a script in bash , $0 expands to bash (or at least, the name of the shell bash was started as). So $(echo $0) expands first to $(echo bash) , which evaluates to bash , which is then identified as the name of the command to run, so a new shell instance is started. You can observe the nesting by running

echo foo; source myscript.sh; echo bar

First you'll see the string foo printed, then you'll be at a prompt in the new shell. Type exit to exit it, and you'll then see bar printed as your original command list finally completes.