CODEWITHSUNDEEP
zsh
balki
balki

Reputation: 27684

How to exec but keeping the same argv0

From zshbuiltins manual,

   exec [ -cl ] [ -a argv0 ] simple command
          Replace  the  current shell with an external command rather than forking.  With -c clear the environment; with -l prepend - to the argv[0] string of the command executed (to simulate a login shell); 
   with -a argv0 set the argv[0] string  of  the  command executed.  See the section `Precommand Modifiers'.

I tried using -a with a simple script but it doesn't seem to work

[balakrishnan@mylap scripts]$ cat printer.sh;echo "-----";cat wrapper.sh
#!/bin/sh
echo $0 $1 $2 $3 $4
-----
#!/bin/zsh
argv0="$0"
exec -a "$argv0" printer.sh
[balakrishnan@mylap scripts]$ wrapper.sh 
printer.sh
[balakrishnan@mylap scripts]$

Since I set wrapper.sh as argv0, I expect that to be printed when printer.sh echos $0. But it still print printer.sh.

Upvotes: 2

Views: 1191

Answers (1)

andrewdotn
andrewdotn

Reputation: 34863

zsh is setting argv[0] correctly, but when /bin/sh runs to interpret the script, it sets $0 to the name of the script being run, ignoring argv[0].

The zsh man page doesn’t explicitly describe this behaviour, but the bash manpage does:

ARGUMENTS

If arguments remain after option processing, and neither the -c nor the -s option has been supplied, the first argument is assumed to be the name of a file containing shell commands. If bash is invoked in this fashion, $0 is set to the name of the file, and the positional parame- ters are set to the remaining arguments.

You can see that argv[0] is being set correctly by running a tiny C program instead of a shell script:

#include <stdio.h>

int main(int argc, char** argv) {
    printf("%s\n", argv[0]);
    return 0;
}

Upvotes: 1

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