Python：使用 ctypes 访问 DLL 函数——按函数 *name* 访问失败
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Python: accessing DLL function using ctypes -- access by function *name* fails
提问by Vit Bernatik
myPythonClient(below) wants to invoke a ringBellfunction (loaded from a DLL using ctypes). However, attempting to access ringBellvia its nameresults in an AttributeError. Why?
myPythonClient（下面）想要调用一个ringBell函数（使用 DLL 从 DLL 加载ctypes）。但是，尝试ringBell通过其名称访问会导致AttributeError. 为什么？
RingBell.hcontains
RingBell.h包含
namespace MyNamespace
{
class MyClass
{
public:
static __declspec(dllexport) int ringBell ( void ) ;
} ;
}
RingBell.cppcontains
RingBell.cpp包含
#include <iostream>
#include "RingBell.h"
namespace MyNamespace
{
int __cdecl MyClass::ringBell ( void )
{
std::cout << "\a" ;
return 0 ;
}
}
myPythonClient.pycontains
myPythonClient.py包含
from ctypes import *
cdll.RingBell[1]() # this invocation works fine
cdll.RingBell.ringBell() # however, this invocation errors out
# AttributeError: function 'ringBell' not found
采纳答案by Vinay Sajip
Perhaps because the C++ name is mangled by the compiler and not exported from the DLL as RingBell. Have you checked that it appears in the exported names exactly like that?
也许是因为 C++ 名称被编译器修改过，而不是从 DLL 中导出为RingBell. 您是否检查过它是否完全像那样出现在导出的名称中？
回答by Alex Martelli
Your C++ compiler is mangling the names of all externally visible objects to reflect (as well as their underlying names) their namespaces, classes, and signatures (that's how overloading becomes possible).
您的 C++ 编译器正在修改所有外部可见对象的名称以反映（以及它们的底层名称）它们的命名空间、类和签名（这就是重载成为可能的方式）。
In order to avoid this mangling, you need an extern "C"on externally visible names that you want to be visible from non-C++ code (and therefore such names cannot be overloaded, nor in C++ standard can they be inline, within namespaces, or within classes, though some C++ compilers extend the standard in some of these directions).
为了避免这种修改，您需要一个extern "C"外部可见的名称，您希望在非 C++ 代码中看到这些名称（因此此类名称不能重载，在 C++ 标准中，它们也不能内联、在名称空间内或在类内，尽管一些 C++ 编译器在其中一些方向上扩展了标准）。
回答by Vit Bernatik
All is working now :) To summarize your posts:
现在一切正常:) 总结你的帖子：
Write DLL in C++:
用 C++ 编写 DLL：
// Header
extern "C"
{ // Name in DLL will be "MyAdd" - but you won't be able to find parameters etc...
__declspec(dllexport) int MyAdd(int a, int b);
}
// Name will be with lot of prefixes but some other info is provided - IMHO better approach
__declspec(dllexport) int MyAdd2(int a, int b);
//.cpp Code
__declspec(dllexport) int MyAdd(int a, int b)
{ return a+b;
}
__declspec(dllexport) int MyAdd2(int a, int b)
{ return a+b;
}
Then you can use program link.exe to see real function name in dll. link.exe is for example in MSVC2010 here:
然后你可以使用程序link.exe查看dll中的真实函数名。link.exe 例如在 MSVC2010 中：
c:\program files\microsoft visual studio 10.0\VC\bin\link.exe
use:
利用：
link /dump /exports yourFileName.dll
you see Something like:
你会看到类似的东西：
ordinal hint RVA name
1 0 00001040 ?MyAdd2@@YAHHH@Z = ?MyAdd2@@YAHHH@Z (int __cdecl MyAdd2(int,int))
2 1 00001030 MyAdd = _MyAdd
Then in python you can import it as:
然后在python中，您可以将其导入为：
import ctypes
mc = ctypes.CDLL('C:\testDll3.dll')
#mc.MyAdd2(1,2) # this Won't Work - name is different in dll
myAdd2 = getattr(mc,"?MyAdd2@@YAHHH@Z") #to find name use: link.exe /dump /exports fileName.dll
print myAdd2(1,2)
#p1 = ctypes.c_int (1) #use rather c types
print mc[1](2,3) # use indexing - can be provided using link.exe
print mc.MyAdd(4,5)
print mc[2](6,7) # use indexing - can be provided using link.exe
