There is a gotcha here if you use: set -o errexit Because *any bash arithmetic expression that evaluates to zero returns the exit code of 1*. So this works

There is a gotcha here if you use: set -o errexit Because *any bash arithmetic expression that evaluates to zero returns the exit code of 1*. So this works: #!/usr/bin/env bash set -o errexit i=0 ((++i)) echo $i # 1 will be printed to stdout But this doesn't: #!/usr/bin/env bash set -o errexit i=0 ((i++)) # terminates here with $? of 1 echo $i The relevant doc is for `let`: let arg [arg ...] Each arg is an arithmetic expression to be evaluated (see ARITHMETIC EVALUATION above). If the last arg evaluates to 0, let returns 1; 0 is returned otherwise. https://www.gnu.org/software/bash/manual/bash.html#index-let https://www.gnu.org/software/bash/manual/bash.html#index-let