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.