subshell

When working with bash scripting, understanding the behavior of local variable assignment and its interaction with subshell exit codes can be crucial for debugging and ensuring your scripts work as expected. Here, we’ll delve into a common scenario that can cause unexpected behavior, helping both new and seasoned scripters gain deeper insights. A: When you use local var=$(cmd) within a function, the exit code of the cmd is indeed executed, but immediately overwritten by the exit code of the local itself. local is a shell builtin that returns an exit status based on its own execution – mostly successful, hence an exit status of 0. This overwrites the exit code of $(cmd), rendering it unobtainable directly after the assignment.