Question

python ctypes 怎么处理函数返回的一般指针

Answer 1

test.c(动态库源代码)

[cpp] view plain copy
// 编译生成动态库: gcc -g -fPIC -shared -o libtest.so test.c

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

typedef struct StructPointerTest
{
char name[20];
int age;
}StructPointerTest, *StructPointer;

StructPointer test() // 返回结构体指针
{
StructPointer p = (StructPointer)malloc(sizeof(StructPointerTest));
strcpy(p->name, "Joe");
p->age = 20;

return p;
}

编译：gcc -g -fPIC -shared -o libtest.so test.c

call.py(python调用C语言生成的动态库):

[python] view plain copy
#!/bin/env python
# coding=UTF-8

from ctypes import *

#python中结构体定义
class StructPointer(Structure):
_fields_ = [("name", c_char * 20), ("age", c_int)]

if __name__ == "__main__":
lib = cdll.LoadLibrary("./libtest.so")
lib.test.restype = POINTER(StructPointer)
p = lib.test()

print "%s: %d" %(p.contents.name, p.contents.age)

最后运行结果：
[plain] view plain copy
[zcm@c_py #112]$make clean
rm -f *.o libtest.so
[zcm@c_py #113]$make
gcc -g -fPIC -shared -o libtest.so test.c
[zcm@c_py #114]$./call.py
Joe: 20
[zcm@c_py #115]$