63

Oh my word I'm a fool. I was simply omitting the second and third arguments when calling the function. Like a fool. Because that's what I am. Original silly question follows:

This seems like it must be a very common thing to do, but I can't find a relevant tutorial, and I'm too ignorant about Numpy and ctypes to figure it out myself.

I have a C function in file ctest.c.

#include <stdio.h>

void cfun(const void * indatav, int rowcount, int colcount, void * outdatav) {
    //void cfun(const double * indata, int rowcount, int colcount, double * outdata) {
    const double * indata = (double *) indatav;
    double * outdata = (double *) outdatav;
    int i;
    puts("Here we go!");
    for (i = 0; i < rowcount * colcount; ++i) {
        outdata[i] = indata[i] * 2;
    }
    puts("Done!");
}

(As you may guess, I originally had the arguments as double * rather than void *, but couldn't figure out what to do on the Python side. I'd certainly love to change them back, but I'm not picky as long as it works.)

I make a shared library out of it. gcc -fPIC -shared -o ctest.so ctest.c

Then in Python, I have a couple numpy arrays, and I'd like to pass them to the C function, one as input and one as output.

indata = numpy.ones((5,6), dtype=numpy.double)
outdata = numpy.zeros((5,6), dtype=numpy.double)
lib = ctypes.cdll.LoadLibrary('./ctest.so')
fun = lib.cfun
# Here comes the fool part.
fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_void_p(outdata.ctypes.data))

print 'indata: %s' % indata
print 'outdata: %s' % outdata

This doesn't report any errors, but prints out

>>> Here we go!
Done!
indata: [[ 1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.]
 [ 1.  1.  1.  1.  1.  1.]]
outdata: [[ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.]]

The outdata array is not modified. And in fact if I call the function again I get a segfault. Which doesn't surprise me -- I really don't know what I'm doing here. Can anyone point me in the right direction?

4

2 回答 2

84

虽然不是对您最初问题的直接回答，但这里有一种更方便的方式来调用您的函数。首先，使 C 函数的原型与在普通 C 中所做的完全一样。由于您不需要单独rowcount，colcount我将它们折叠成一个size参数：

void cfun(const double *indatav, size_t size, double *outdatav) 
{
    size_t i;
    for (i = 0; i < size; ++i)
        outdatav[i] = indatav[i] * 2.0;
}

现在按以下方式定义 ctypes 原型：

import ctypes
from numpy.ctypeslib import ndpointer
lib = ctypes.cdll.LoadLibrary("./ctest.so")
fun = lib.cfun
fun.restype = None
fun.argtypes = [ndpointer(ctypes.c_double, flags="C_CONTIGUOUS"),
                ctypes.c_size_t,
                ndpointer(ctypes.c_double, flags="C_CONTIGUOUS")]

现在，调用您的函数将非常方便：

indata = numpy.ones((5,6))
outdata = numpy.empty((5,6))
fun(indata, indata.size, outdata)

您还可以定义一个包装器以使其更加方便：

def wrap_fun(indata, outdata):
    assert indata.size == outdata.size
    fun(indata, indata.size, outdata)
于 2011-05-03T10:16:43.460 回答
21

只需将所有四个参数传递给 C 函数。从以下位置更改您的 Python 代码：

fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_void_p(outdata.ctypes.data))

至：

fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_int(5), ctypes.c_int(6),
    ctypes.c_void_p(outdata.ctypes.data))
于 2011-05-02T23:21:17.053 回答