Re: 'exec -a' and $0 substitution

[Lawrence Velázquez]

On Wed, Jan 25, 2023, at 6:37 PM, Sergei Trofimovich wrote:
> I fear it's a side-effect of the way 'bash' gets executed via shebang by
> kernel. But maybe not? Somehow direct script execution still manages to
> preserve script's name. Is it an intended behaviour that could not be
> easily changed? Or it's a bug?

It's not a bug.

https://www.gnu.org/software/bash/manual/html_node/Special-Parameters.html#index-0

        0
                ($0)  Expands to the name of the shell or shell
                script.  This is set at shell initialization.  If
                Bash is invoked with a file of commands (see Shell
                Scripts), $0 is set to the name of that file.  If
                Bash is started with the -c option (see Invoking
                Bash), then $0 is set to the first argument after
                the string to be executed, if one is present.
                Otherwise, it is set to the filename used to invoke
                Bash, as given by argument zero.

Whenever bash is executed with a script file, it sets $0 to the
name of that file.  It only uses caller-provided values for $0 when
a script file is not provided.

        $ (exec -a foo bash <<<'echo "$0"')
        foo
        $ (exec -a foo bash -s bar baz quux <<<'echo "$0"')
        foo
        $ (exec -a foo bash -c 'echo "$0"')
        foo
        $ (exec -a foo bash -c 'echo "$0"' bar baz quux)
        bar

This behavior is standardized in POSIX [*] and is not unique to
bash (I omitted ksh93, which actually does produce "foo" here).

        $ (exec -a foo zsh <(echo 'echo "$0"'))
        /dev/fd/63
        $ (exec -a foo dash <(echo 'echo "$0"'))
        /dev/fd/63
        $ (exec -a foo yash <(echo 'echo "$0"'))
        /dev/fd/63
        $ (exec -a foo mksh <(echo 'echo "$0"'))
        /dev/fd/63

[*]: 
https://pubs.opengroup.org/onlinepubs/9699919799/utilities/sh.html#tag_20_117_05

-- 
vq