当使用ctypes与Python和C ++时的Seg故障 [英] Seg fault when using ctypes with Python and C++

问题描述

我在Linux机器上尝试使用ctypes从Python 2.7调用C ++库。目前，我只是玩玩具例子。我阅读并能够复制此线程中提供的示例：从python调用C / C ++？。注意，C ++类函数不需要输入。

我试图扩展这个例子如下。

foo.cpp

  #include< iostream> 
 #include< string> 
 using namespace std; 
 
 class Foo {
 public：
 Foo（string name）; 
 void bar（void）; 
 private：
 string myName; 
}; 
 Foo :: Foo（string name）{
 std :: cout<< Entered constructor in C ++< std :: endl; 
 myName = name; 
} 
 void Foo :: bar（void）{
 std :: cout< 你好，这是< myName<< std :: endl; 
} 
 
 externC{
 Foo * Foo_new（string name）{return new Foo（name）;} 
 void Foo_bar（Foo * foo）{ foo> bar（）;} 
} 
  

c $ c> g ++ -c -fPIC foo.cpp -o foo.o 然后 g ++ -shared -Wl，-soname，libfoo.so -o libfoo.so foo

fooWrapper.py

 来自ctypes import cdll，c_char_p 
 lib = cdll.LoadLibrary（'./ libfoo.so'）
 
 class Foo（object）：
 def __init __（self，name）：
 self.name = name 
 self.obj = lib。 Foo_new（c_char_p（name））
 def bar（self）：
 lib.Foo_bar（self.obj）
  

因为我传递一个字符串到构造函数，我的印象是，我需要使用 c_char_p 。也许这是错误的。

当我在Python 2.7中执行代码时，我得到一个分段错误，

  Python 2.7.2（默认，2012年2月27日，18:28:19）
 [GCC 4.1.2 20080704（Red Hat 4.1.2-48）] on linux2 
键入帮助，版权，信用或许可证了解更多信息。 
>>>> import fooWrapper as f 
>>>> foo = f.Foo（'somename'）
分段错误
  

C ++构造函数。我有同样的问题，无论是否在Python代码的构造函数方法中包含 self.name = name 行。

我做错了什么？感谢。

解决方案

在您的 externC c $ c> Foo_new 获取一个 std :: string 参数，但是对于可以从Python调用的代码， C函数，其中包括接收字符串的字符指针。特别是你实际上告诉解释器将你的字符串转换为一个字符指针（这是 c_char_p 调用）。

I am trying to get up and running with using ctypes to call C++ libraries from Python 2.7 on a Linux machine. For the moment, I'm just trying toy examples. I read and was able to replicate the example given in this thread: Calling C/C++ from python?. Note that the C++ class functions take no inputs.

I'm trying to extend this example as follows.

foo.cpp

#include <iostream>
#include <string>
using namespace std;

class Foo {
public:
  Foo(string name);
  void bar(void);
private:
  string myName;
};
Foo::Foo(string name) {
  std::cout << "Entered constructor in C++" << std::endl;
  myName = name;
}
void Foo::bar(void) {
  std::cout << "Hello, this is " << myName << std::endl;
}

extern "C" {
  Foo* Foo_new(string name) {return new Foo(name);}
  void Foo_bar(Foo* foo) {foo->bar();}
}

The code compiles cleanly using g++ -c -fPIC foo.cpp -o foo.o and then g++ -shared -Wl,-soname,libfoo.so -o libfoo.so foo.o.

I then use the following wrapper,

fooWrapper.py

from ctypes import cdll, c_char_p
lib = cdll.LoadLibrary('./libfoo.so')

class Foo(object):
    def __init__(self,name):
        self.name = name
        self.obj = lib.Foo_new(c_char_p(name))
    def bar(self):
        lib.Foo_bar(self.obj)

Because I'm passing a string to the constructor, I'm under the impression that I need to cast it using c_char_p. Maybe this is wrong.

When I execute the code in Python 2.7, I get a segmentation fault,

Python 2.7.2 (default, Feb 27 2012, 18:28:19) 
[GCC 4.1.2 20080704 (Red Hat 4.1.2-48)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import fooWrapper as f
>>> foo = f.Foo('somename')
Segmentation fault

So I never get into the C++ constructor function. I have the same problem whether or not I include the self.name = name line in the constructor method in the Python code.

What am I doing wrong? Thanks.

解决方案

In your extern "C" you declare Foo_new to take a std::string argument, however for the code to be callable from Python you have to make it a true C function, which include receiving strings a character pointers. Especially as you actually tells the interpreter to convert your string to a character pointer (that's what the c_char_p call does).

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