9

以下のように、一次元配列を python に渡すことができました。そして、ctypes、numpyを使用して、c++のダブルポインター配列をpythonに渡すことができるかどうか疑問に思います。

test.cpp:

#include <stdio.h>
extern "C" void cfun(const void * indatav, int rowcount, int colcount, void * outdatav);

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] * 4;
    }
    puts("Done!");
}

test.py:

import numpy
import ctypes

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))

fun(ctypes.c_void_p(indata.ctypes.data), ctypes.c_int(5), ctypes.c_int(6),
    ctypes.c_void_p(outdata.ctypes.data))


print 'indata: %s' % indata
print 'outdata: %s' % outdata
4

1 に答える 1

8

ここに方法があります。double** で numpy を使用する良い方法がわかりませんでした。

test.cpp (Windows)

#include <stdio.h>

extern "C" __declspec(dllexport) void cfun(const double ** indata, int rowcount, int colcount, double ** outdata) {
    for (int i = 0; i < rowcount; ++i) {
        for (int j = 0; j < colcount; ++j) {
            outdata[i][j] = indata[i][j] * 4;
        }
    }
}

test.py

import numpy
import ctypes

# Allocate array of double*
indata = (ctypes.POINTER(ctypes.c_double) * 5)()
for i in range(5):
    # Allocate arrays of double
    indata[i] = (ctypes.c_double * 6)()
    for j in range(6):
        indata[i][j] = 1.0

outdata = (ctypes.POINTER(ctypes.c_double) * 5)()
for i in range(5):
    outdata[i] = (ctypes.c_double * 6)()
    for j in range(6):
        outdata[i][j] = 1.0

lib = ctypes.cdll.LoadLibrary('test')
fun = lib.cfun

def dump(a,rows,cols):
    for i in range(rows):
        for j in range(cols):
            print a[i][j],
        print

dump(indata,5,6)
fun(ctypes.byref(indata),5,6,ctypes.byref(outdata))
dump(outdata,5,6)

出力

1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
4.0 4.0 4.0 4.0 4.0 4.0
于 2012-01-09T03:55:39.170 に答える