On Nov 8, 3:35 pm, Hans Mulder <han...@xs4all.nl> wrote: > > Perhaps this example better demonstrates what is going on: > > >>>> p = subprocess.Popen(['echo one $0 three $1 five', 'two', 'four'], > > ... shell=True) > > one two three four five > > Maybe I'm thick, but I still don't understand. If I run a shell script, > then the name of the script is in $0 and the first positional arguments > is in $1, similar to how Python sets up sys.argv. > > But in this case the first positional argument is in $0. Why is that?
It's just a quirk in the way the shell handles the -c option. The syntax for the shell invocation boils down to something like this: sh [-c command_string] command_name arg1 arg2 arg3 ... Without the -c option, sh runs the file indicated by command_name, setting $0 to command_name, $1 to arg1, $2 to arg2, etc. With the -c option, it does the same thing; it just runs the command_string instead of a file pointed to by command_name. The latter still conceptually exists as an argument, however, which is why it still gets stored in $0 instead of $1. We could argue about whether this approach is correct or not, but it's what the shell does, and that's not likely to change. Cheers, Ian -- http://mail.python.org/mailman/listinfo/python-list
