[Tutor] os.execl() args?

[Tadahiko 'kiko' Uehara]

Thank you for replying, David!

In my enviroment (Linux,tcsh,python2.2), I don't see path name.. :(

###
jem:/home/kiko/docs/python> python args.py lots of arguments here
Args were:
args.py
lots
of
arguments
here
###

Basically, My understand for arguments are (optional) parameters which passed to command. 
	ie. 'xv foo.png'  (xv is command and foo.png is an argument)
	execl('xv','','foo.png')

	or 'mv -r foo foo2' (mv for command, -r , foo , foo2 are arguments)

Is this correct? 

Thanks in advance,
-kiko



On Sat, 10 May 2003 15:45:53 -0400
"David Broadwell" <dbroadwell@mindspring.com> wrote:

> <questioner wrote>
> 2) os.execl(render,'',queue) , I had to give arg0 blank. I found the
> following explanation in the doc...
> 
> "For the C programmer, this is the argv[0] passed to a program's main(). For
> example, "os.execv('/bin/echo', ['foo', 'bar'])" will only print "bar" on
> standard output; "foo"will seem to be ignored."
> 
>  I'm not quite sure what it says.. Though, give argv0 to nothing'' is
> correct?
> 
> 
> Well, I'm green to, so I'll answer with code.
> 
> <file: args.py>
> from sys import argv
> 
> print "Args were:"
> for item in argv:
> 	print item
> </file: args.py>
> 
> $p$g > args.py lots of arguments here
> Args were:
> D:\mongoose\My Documents\My Code\python\args.py
> lots
> of
> arguments
> here
> 
> ---
> 
> sys.argv[0] is the path to the script ... And it's interesting to note that
> if you "" a block of args they end up as one arg.
> 
> Sorry, can't really answer the rest.
> 
> --
> 
> David Broadwell (Amar Shadowpaws)
>