Why must path be in args in os.execv?

[Holger Jannsen]

From: Holger Jannsen <holger at phoenix-edv.netzservice.de>

Thanks, Jeff.

That's a good answer.

Ciao,
Holger

Jeff Epler schrieb:
> 
> Holger,
> 
> On Unix, read the manpage for execve and friends.  You will see that the
> first argument is the path to the executable, and the subsequent arguments
> describe the arguments (including argv[0]) and optionally the environment.
> By convention, argv[0] is the basename of the executable when the
> executable is on the PATH, and the full path otherwise.  (Another unix
> convention is that when the first letter of argv[0] is "-", the shell is a
> login shell)
> 
> DOS doesn't really have execve.  It's possible that the imitation is less
> perfect than it could be.  Instead, it may do the equivalent of entering
> the string
>         string.join(argv, " ")
> in the shell, i.e., ignoring the path argument.
> 
> This looks like a shortcoming in the DOS (windows 9x, NT) implementation of
> exec*, in a way that makes it work differently from the Unix it attempts to
> imitate.  This wouldn't surprise me, since I recently discovered that NT's
> spawnl takes a unix-like array of arguments (const char *argv[]) but when
> any element contains a space, it is split into two arguments in the child
> process.  So, I had to break some code that had worked fine on Unix to
> generate argv, by quoting each argument (something that would *not* work on
> Unix) ..
> 
> Bad enough that Windows wants to copy Unix, they have to consistently get
> it subtly wrong too.
> 
> Jeff