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Why must path be in args in os.execv?

Jeff Epler jepler at inetnebr.com
Thu Jul 1 11:26:52 EDT 1999 
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
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