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I'm trying to make a shell for the desktop environment I'm designing which is just a configured Zsh, and I'm confused about how to pass along whether it's a login shell or not.

Here's my script:

#!/bin/sh
case "$0" in
-*) ZDOTDIR=/etc/tiles/zsh exec zsh -l "$@";;
*) ZDOTDIR=/etc/tiles/zsh exec zsh "$@";;
esac

I have installed it to /bin/tiles-shell, added that path to /etc/shells, and made it my shell with chsh.

The problem is it's never a login shell because $0 seems to always be /bin/tiles-shell. I thought it would start with - when it should be a login shell?

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  • 1
    If it's for zsh why have you indicated it's a sh script (top line)? Commented Nov 7, 2022 at 13:19
  • @roaima A wrapper can use whatever works before replacing itself with the actual tool. Commented Nov 7, 2022 at 13:22
  • @KamilMaciorowski indeed it can, but I wanted to check it's what was intended Commented Nov 7, 2022 at 13:23
  • 1
    $0 would only start with - I think? So you'd want something more like case $0 in -*) Commented Nov 7, 2022 at 13:23
  • and shouldn't zsh recognize the - at the start of $0 itself anyway? is the wrapper really needed for that? Commented Nov 7, 2022 at 13:29

2 Answers 2

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4

The shebang mechanism wipes the old value of argv[0]: the kernel replaces it by the path to the interpreter.

The shell does not expose the value of argv[0] as $0 when running a script from a file. $0 is the name of the script file. I don't think zsh exposes its argv[0] through any other interface.

Demonstration script (for Linux):

#!/bin/zsh
echo "argv[0]=$(</proc/$$/cmdline tr \\0 \\n | head -n 1)"
echo "\$0=$0"

Output:

% ARGV0=foo zsh -c "$(cat a)"
argv[0]=foo
$0=foo
% ARGV0=foo zsh ./a
argv[0]=foo
$0=./a
% ARGV0=foo ./a
argv[0]=/bin/zsh
$0=./a

If you really want your login shell to be a script, make it a script that is dedicated to being a login shell.

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1

After Giles' answer I made it a binary and it works.

#include <stdlib.h>
#include <unistd.h>

int main(int argc, char *argv[]) {
    putenv("ZDOTDIR=/etc/tiles/zsh");
    execv("/usr/bin/zsh", argv);
}
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