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 OperationSPTobj = pointer(StructPointerTest(1, 2))print SPTobjprint 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):passStructPointer._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 therestypeattributeof the function object.
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