[Numpy-discussion] Getting C-function pointers from Python to C

Nadav Horesh nadavh@visionsense....
Tue Apr 10 10:25:32 CDT 2012


Sorry for being slow.
There is (I think) a related question I raised on the skimage list:
I have a cython function that calls a C callback function in a loop (one call for each pixel in an image). The C function in compiled in a different shared library (a simple C library, not a python module). I would like a python script to get the address of the C function and pass it on to the cython function as the pointer for the callback function.

As I understand Travis' isue starts ones the callback address is obtained, but, is there a direct method to retrieve the address from the shared library?

   Nadav.
________________________________________
From: numpy-discussion-bounces@scipy.org [numpy-discussion-bounces@scipy.org] On Behalf Of Travis Oliphant [teoliphant@gmail.com]
Sent: 10 April 2012 03:11
To: Discussion of Numerical Python
Subject: [Numpy-discussion] Getting C-function pointers from Python to C

Hi all,

Some of you are aware of Numba.   Numba allows you to create the equivalent of C-function's dynamically from Python.   One purpose of this system is to allow NumPy to take these functions and use them in operations like ufuncs, generalized ufuncs, file-reading, fancy-indexing, and so forth.  There are actually many use-cases that one can imagine for such things.

One question is how do you pass this function pointer to the C-side.    On the Python side, Numba allows you to get the raw integer address of the equivalent C-function pointer that it just created out of the Python code.    One can think of this as a 32- or 64-bit integer that you can cast to a C-function pointer.

Now, how should this C-function pointer be passed from Python to NumPy?   One approach is just to pass it as an integer --- in other words have an API in C that accepts an integer as the first argument that the internal function interprets as a C-function pointer.

This is essentially what ctypes does when creating a ctypes function pointer out of:

  func = ctypes.CFUNCTYPE(restype, *argtypes)(integer)

Of course the problem with this is that you can easily hand it integers which don't make sense and which will cause a segfault when control is passed to this "function"

We could also piggy-back on-top of Ctypes and assume that a ctypes function-pointer object is passed in.   This allows some error-checking at least and also has the benefit that one could use ctypes to access a c-function library where these functions were defined. I'm leaning towards this approach.

Now, the issue is how to get the C-function pointer (that npy_intp integer) back and hand it off internally.   Unfortunately, ctypes does not make it very easy to get this address (that I can see).    There is no ctypes C-API, for example.    There are two potential options:

        1) Create an API for such Ctypes function pointers in NumPy and use the ctypes object structure.  If ctypes were to ever change it's object structure we would have to adapt this API.

        Something like this is what is envisioned here:

             typedef struct {
                        PyObject_HEAD
                        char *b_ptr;
             } _cfuncptr_object;

        then the function pointer is:

            (*((void **)(((_sp_cfuncptr_object *)(obj))->b_ptr)))

        which could be wrapped-up into a nice little NumPy C-API call like

        void * Npy_ctypes_funcptr(obj)


        2) Use the Python API of ctypes to do the same thing.   This has the advantage of not needing to mirror the simple _cfuncptr_object structure in NumPy but it is *much* slower to get the address.   It basically does the equivalent of

        ctypes.cast(obj, ctypes.c_void_p).value


        There is working code for this in the ctypes_callback branch of my scipy fork on github.


I would like to propose two things:

        * creating a Npy_ctypes_funcptr(obj) function in the C-API of NumPy and
        * implement it with the simple pointer dereference above (option #1)


Thoughts?

-Travis







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