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I am trying to beef up my notify script. The way the script works is that I put it behind a long running shell command and then all sorts of notifications get invoked after the long running script finished. For example: sleep 100; my_notify It would be nice to get the exit code of the long running script. The problem is that calling my_notify creates a new process that does not have access to the $? variable. Compare: ~ $: ls nonexisting_file; echo "exit code: $?"; echo "PPID: $PPID" ls: nonexisting_file: No such file or directory exit code: 1 PPID: 6203 vs. ~ $: ls nonexisting_file; my_notify ls: nonexisting_file: No such file or directory exit code: 0 PPID: 6205 The my_notify script has the following in it: #!/bin/sh echo "exit code: $?" echo "PPID: $PPID" I am looking for a way to get the exit code of the previous command without changing the structure of the command too much. I am aware of the fact that if I change it to work more like time , e.g. my_notify longrunning_command... my problem would be solved, but I actually like that I can tack it at the end of a command and I fear complications of this second solution. Can this be done or is it fundamentally incompatible with the way that shells work? My shell is Z shell ( zsh ), but I would like it to work with Bash as well.

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