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Question about bash behavior with non-interactive mode

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Thomas Dziedzic

I basically need to understand the following behavior:

$ echo $SHELL
/bin/bash
$ bash -c echo $SHELL

bash -c echo $SHELL produces a blank output, and this confuses me because I would expect the SHELL to be set. This is true for $BASH_VERSION as well.

Can someone explain this to me?

Thanks for your time!

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steeldriver

From man bash:

   -c        If the -c option is present, then commands are read from  the
             first non-option argument command_string.  If there are argu‐
             ments after the command_string, the  first  argument  is  as‐
             signed  to $0 and any remaining arguments are assigned to the
             positional parameters.

In bash -c echo $SHELL, the "first non-option argument" is echo, while $SHELL is expanded by your current interactive shell and passed to bash as positional parameter $0. Since echo has no arguments itself, bash -c echo prints an empty string. Meanwhile, $0 is set to /bin/bash but ignored.

Depending on what you are trying to do, you can either do

bash -c 'echo $SHELL'

which passes echo $SHELL as command_string to the non-interactive bash shell (outputing the value of the non-interactive shell's $SHELL), or

bash -c 'echo $0' $SHELL

which passes (the expansion of) the interactive shell's $SHELL to the non-interactive bash shell as $0, then echoes its value.

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