求python 传参与接收代码（ctypes）！

Question

c:
char * a(char *b)
{
return b;
}

注意char *不是字符串，里面有一堆\0

Answer 1

Python是解释性语言, 底层就是用c实现的, 所以用python调用C是很容易的, 下面就总结一下各种调用的方法, 给出例子, 所有例子都在ubuntu9.10, python2.6下试过1. Python 调用 C (base)想在python中调用c函数, 如这儿的fact#include <Python.h>int fact(int n){if (n <= 1)return 1;elsereturn n * fact(n - 1);}PyObject* wrap_fact(PyObject* self, PyObject* args){int n, result;if (! PyArg_ParseTuple(args, "i:fact", &n))return NULL;result = fact(n);return Py_BuildValue("i", result);}static PyMethodDef exampleMethods[] ={{"fact", wrap_fact, METH_VARARGS, "Caculate N!"},{NULL, NULL}};void initexample(){PyObject* m;m = Py_InitModule("example", exampleMethods);}把这段代码存为wrapper.c, 编成so库,gcc -fPIC wrapper.c -o example.so -shared -I/usr/include/python2.6 -I/usr/lib/python2.6/config然后在有此so库的目录, 进入python, 可以如下使用import exampleexample.fact(4)2. Python 调用 C++ (base)在python中调用C++类成员函数, 如下调用TestFact类中的fact函数,#include <Python.h>class TestFact{public:TestFact(){};~TestFact(){};int fact(int n);};int TestFact::fact(int n){if (n <= 1)return 1;elsereturn n * (n - 1);}int fact(int n){TestFact t;return t.fact(n);}PyObject* wrap_fact(PyObject* self, PyObject* args){int n, result;if (! PyArg_ParseTuple(args, "i:fact", &n))return NULL;result = fact(n);return Py_BuildValue("i", result);}static PyMethodDef exampleMethods[] ={{"fact", wrap_fact, METH_VARARGS, "Caculate N!"},{NULL, NULL}};extern "C" //不加会导致找不到initexamplevoid initexample(){PyObject* m;m = Py_InitModule("example", exampleMethods);}把这段代码存为wrapper.cpp, 编成so库,g++ -fPIC wrapper.cpp -o example.so -shared -I/usr/include/python2.6 -I/usr/lib/python2.6/config然后在有此so库的目录, 进入python, 可以如下使用import exampleexample.fact(4)3. Python 调用 C++ (Boost.Python)Boost库是非常强大的库, 其中的python库可以用来封装c++被python调用, 功能比较强大, 不但可以封装函数还能封装类, 类成员.http://dev.gameres.com/Program/Abstract/Building%20Hybrid%20Systems%20with%20Boost_Python.CHN.by.JERRY.htm首先在ubuntu下安装boost.python, apt-get install libboost-python-dev#include <boost/python.hpp>char const* greet(){return "hello, world";}BOOST_PYTHON_MODULE(hello){using namespace boost::python;def("greet", greet);}把代码存为hello.cpp, 编译成so库g++ hello.cpp -o hello.so -shared -I/usr/include/python2.5 -I/usr/lib/python2.5/config -lboost_python-gcc42-mt-1_34_1此处python路径设为你的python路径, 并且必须加-lboost_python-gcc42-mt-1_34_1, 这个库名不一定是这个, 去/user/lib查然后在有此so库的目录, 进入python, 可以如下使用>>> import hello>>> hello.greet()'hello, world'4. python 调用 c++ (ctypes)ctypes is an advanced ffi (Foreign Function Interface) package for Python 2.3 and higher. In Python 2.5 it is already included.ctypes allows to call functions in dlls/shared libraries and has extensive facilities to create, access and manipulate simple and complicated C data types in Python - in other words: wrap libraries in pure Python. It is even possible to implement C callback functions in pure Python.http://python.net/crew/theller/ctypes/#include <Python.h>class TestFact{public:TestFact(){};~TestFact(){};int fact(int n);};int TestFact::fact(int n){if (n <= 1)return 1;elsereturn n * (n - 1);}extern "C"int fact(int n){TestFact t;return t.fact(n);}将代码存为wrapper.cpp不用写python接口封装, 直接编译成so库,g++ -fPIC wrapper.cpp -o example.so -shared -I/usr/include/python2.6 -I/usr/lib/python2.6/config进入python, 可以如下使用>>> import ctypes>>> pdll = ctypes.CDLL('/home/ubuntu/tmp/example.so')>>> pdll.fact(4)12&nbsp;

Answer 2

我觉得会受到限制的，因为c_char_p是遵循c字符串标准的，会以NULL为结束。下面的代码只输出hello,也许真要传递内嵌NULL的，只能靠编写python扩展了，也很简单的，用swig。from ctypes import *import struct&nbsp;example=cdll.LoadLibrary("example.dll")s=create_string_buffer('hello\x00world')example.test.restype=c_char_pexample.test.argtypes = [c_char_p]r=example.test(s) #("hello\x00world")print r

Answer 3

如果有一堆\0,你怎么知道这个指针指向的内容什么时候结束？比如应该读取10个字节，还是100个字节。最起码要加一个标示指针所指内容长度的参数。