使用ctypes将malloc分配的数组从C返回到Python
我想使用一些C语言代码,这些代码可以返回多个大小不确定的数组。因为有多个数组,所以我觉得需要用到指针传递,但我不太确定怎么把这和用来设置数组的malloc结合起来。
这是一些示例的C代码:
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
//gcc -fPIC -shared -o array_test_c.so array_test_c.c
void return_array(int * outdata1v, int * outdata2v) {
int i;
int N = 10;
int * mydatav2, * mydatav3;
mydatav2 = (int *) malloc(sizeof(int) * N);
mydatav3 = (int *) malloc(sizeof(int) * N);
for (i = 0; i<N; i++){
mydatav2[i] = i;
mydatav3[i] = i*2;
}
//this doesn't work which makes sense
outdata1v = mydatav2;
outdata2v = mydatav3;
}
我想把它和Python连接起来,类似这样(但这个方法不行):
import os
import ctypes
#for c interface
test_module = ctypes.cdll.LoadLibrary(
os.path.join(os.path.dirname(__file__), './array_test_c.so'))
outdata1 = (ctypes.c_int * 0)()
outdata2 = (ctypes.c_int * 0)()
test_module.return_array(outdata1, outdata2)
outdata1 = (ctypes.c_int*10).from_address(ctypes.addressof(outdata1))
print "out", outdata1[-1], outdata1, outdata2
这个方法不行,我总是无法打印出20。有没有什么想法?
2 个回答
2
你需要用到指向指针的指针:
void return_array(int **outdata1v, int **outdata2v) {
int i;
int N = 10;
int * mydatav2, * mydatav3;
mydatav2 = (int *) malloc(sizeof(int) * N);
mydatav3 = (int *) malloc(sizeof(int) * N);
for (i = 0; i<N; i++){
mydatav2[i] = i;
mydatav3[i] = i*2;
}
*outdata1v = mydatav2;
*outdata2v = mydatav3;
}
我对Python的部分不太了解。
4
test.c
#include <stdlib.h>
#define N 10
void test(int *size, int **out1, int **out2) {
int i;
int *data1, *data2;
data1 = (int *)malloc(sizeof(int) * N);
data2 = (int *)malloc(sizeof(int) * N);
for (i = 0; i < N; i++){
data1[i] = i;
data2[i] = i * 2;
}
*size = N;
*out1 = data1;
*out2 = data2;
}
test.py
from ctypes import CDLL, POINTER, c_int, byref
dll = CDLL('test.so')
data1 = POINTER(c_int)()
data2 = POINTER(c_int)()
size = c_int()
dll.test(byref(size), byref(data1), byref(data2))
for i in range(size.value):
print i, data1[i], data2[i]
补充说明:你应该考虑提供一个函数来释放用malloc分配的数据。这样你就可以像这样使用,比如 dll.cleanup(data1, data2)