Why must path be in args in os.execv?

[Jeff Epler]

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