我正在寻找使用一些C代码来返回未知大小的多个数组。因为有多个数组,所以我认为我需要使用传入的指针,而且我不确定如何将其与用于设置数组的malloc结合使用。
这是一些代表性的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;
}
而且我正在尝试通过以下方式将其连接到Python(这不起作用):
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
这是行不通的,而且我永远也无法打印20张。有任何想法吗?
测试
#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]
编辑:您应该考虑提供释放malloc数据的功能。这样就可以了,例如dll.cleanup(data1, data2)
本文收集自互联网,转载请注明来源。
如有侵权,请联系[email protected] 删除。
我来说两句