Thread: _spawnl() - Bug or sadistic "feature"?

I just had a very unpleasant experience when trying to use the _spawnl() function...

The MSDN says the syntax is:

Code:

intptr_t _spawnl(
   int mode,
      const char *cmdname,
   const char *arg0,
   const char *arg1,
   ... const char *argn,
   NULL 
);

So I ran it as:

Code:

_spawnl( _P_NOWAIT, "UserExit.bat", "ChangePassword", "User1", "Password", NULL );

My UserExit.bat file calls ChangePassword.bat (depending on the parameters passed)... but I noticed that UserExit.bat was never getting called, but somehow ChangePassword.bat was called. For almost a day I thought I was going crazy. Then after looking at some examples that use _spawnl() I noticed that they always pass the same value for the 2nd & 3rd parameters. That's when I realized that the arg0 in the MSDN syntax relates to the argv[0] that gets passed to programs (i.e. the program name), not the first real parameter.
Obviously when I passed "UserExit.bat" twice everything was working fine, but what I found so strange is that it was calling "ChangePassword.bat" (i.e. using the arg0 parameter as the program to run) instead of "UserExit.bat" which is listed as cmdname. Is this a bug in _spawnl()? Why would it run the arg0 parameter, and if that's what it actually does, then what's the point of even specifying a cmdname parameter?

When I looked it up at MSDN, it said that cmdname was "Path of the file to be executed." I agree that that is an extremely misleading name.

Originally Posted by Salem

But it is consistent with the exec() series of functions which originate from Unix-land.

Yeah, I noticed that. Except I believe the spawn functions do a fork() + exec().
I've never used spawn or exec before (at least not from scratch), so this problem really surprised the hell out of me.

So if arg0 is actually the program that _spawnl() will run, then what does it use cmdname for?