Python/C++ interface

[idfx]

In article <3A79D28B.1BEB46A0 at cybermesa.com>,
  Jay O'Connor <joconnor at cybermesa.com> wrote:
> All,
>
> I was doing some testing with the Python->C++ interface and ran into
> something odd.
>
> The example shows calling a C function from python with
>
> 	import spam
> 	spam.system ("ls -l")
>
> and on the C++ side... the signature reads
> 	PyObject* spam_system(PyObject *self, PyObject * args) {
> 		...
> 	}
>
> Now.  On the C++ side, args is considered, and parsed, as a Tuple.
> However, what I found was that a Tuple was not being passed, just a
> String.  The only time a Tuple was passed was if there was more than
one
> argument.  The whole argument list would be converted to a Tuple,
then.
>
> Is this an error in the documentation?  Or has the passing method
> changed between 1.5.x and 2.0 (I'm using 2.0) and the documentation is
> just out of date?
>
> Also,  is there a good way of printing the name of the type of a
> PyObject?

     I may be wrong, or over-simplifying, but I think you've gotten
confused by the terms, understandably.  Yes, args is read into a tuple,
always, but if there's only one argument, then it becomes a tuple of
length 1.

ie:    tpl = ("yourstring",)

that way, you can refer to as many arguments as there are,

( numargs=len(tpl) )

instead of worrying about special cases for 1 or 0 arguments.
Your args is still a string, it's just that it's a string that happens
to be an element of a tuple.


   OTOH, I may be totally mis-reading your problem.  The other
possibility is that, for a system call, (e.g. 'ls -l'), C++ makes a
string of the argument(s) to preserve the command you are passing to
the system, which should be a string anyway, or so I'm told.


    As for the printing of type names, this is something I had trouble
with, too.  Eventually, I came up with this, although I'm sure there
are more elegant ways to address the problem:

def printype( obj ):
	naym = `type(obj)`       #    string it,
	i = naym.find("'")
	naym = naym[(i+1):]      #    trim it,
	i = naym.find("'")
	naym = naym[:i]
	return naym              #    return it!   (as a string)

    Hey, what can I say? It works. Usually.    ; )

Hope that helps.

--
yrs,
in_sanity,

idfx


Sent via Deja.com
http://www.deja.com/