Python Ctypes 结构体指针处理(函数参数，函数返回)

C函数需要传递结构体指针是常事，但是和Python交互就有点麻烦事了，经过研究也可以了。

<结构体指针作为函数参数>

来看下C测试例子:

#include <stdio.h>
typedef struct StructPointerTest* StructPointer;
struct StructPointerTest{
        int x;
        int y;
};
void test(StructPointer p) {
        p->x = 101;
        p->y = 201;
}

这里test里面需要传入结构体指针,函数中的实现很简单，就是改变x 和 y 的值这个函数将被python调用。

使用Python调用时，需要模拟申明个结构体(class):

from ctypes import *
class StructPointerTest(Structure): 
    _fields_ =[('x', c_int),
               ('y', c_int)]

Usage:

##Structure Pointer Operation
SPTobj = pointer(StructPointerTest(1, 2))
print SPTobj
print SPTobj.contents.x 
print SPTobj.contents.y

<函数返回结构体指针>

C函数测试例子改成如下:

StructPointer test() {
        StructPointer p = (StructPointer)malloc(sizeof(struct StructPointerTest));
        p->x = 101;
        p->y = 201;
        return p;
}

Python程序处理如下:

from ctypes import *
class StructPointer(Structure):
        pass

StructPointer._fields_=[('x', c_int),
                        ('y', c_int),
                        ('next', POINTER(StructPointer))]


lib = cdll.LoadLibrary('./StructPointer.so')
lib.test.restype = POINTER(StructPointer)

p = lib.test()
print p.contents.x

关于resttype可以参见 Tutorial :By default functions are assumed to return the Cinttype. Other return types can be specified by setting therestypeattribute of the function object.