Normal shell vs subshell vs "bash -c" last command exit code

Question

I am trying to get the exit code of the last command chained with && and || correctly.

I encountered a strange behavior that I cannot explain. Please help.

This is my exit-code.sh test script:

#!/bin/bash echo "running exit-code with $1" exit $1 

This works as expected:

$ ./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $? running exit-code with 1 last command exit code: 1 

The same with () to run it in a subshell:

$ (./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $?) running exit-code with 1 last exit code: 1 

But:

$ /bin/bash -c "./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $?" running exit-code with 1 last exit code: 0 

Why do I get exit code 0 here?

Answer 1

Why do I get exit code 0 here?

The reason is that with the double quotes, parameter expansion (of the special parameter, $?) is carried out before passing arguments to the new Bash process. If you turn on debug and verbose mode, you can see this clearly:

$ set -xv $ bash -c "(./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $?)" bash -c "(./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $?)" + bash -c '(./exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: 0)' running exit-code with 1 last exit code: 0 $ set +xv 

In this case $? is set to 0 because the previous command (set -xv in this example) executed successfully.

Answer 2

Quotes

Just use single ' quotes

/bin/bash -c './exit-code.sh 1 && ./exit-code.sh 2 && ./exit-code.sh 3 || echo last exit code: $?' running exit-code with 1 last exit code: 1