CODEWITHSUNDEEP
c++dllexport
aj3423
aj3423

Reputation: 2541

export a c++ function with it's raw function name

I'm writing a dll, which exports a function 

extern "C"
__declspec(dllexport)
std::list<string> create() {
return std::list<string>();
}

the compiler complains:

error C2526: 'create' : C linkage function cannot return C++ class 'std::list<_Ty>'

if I remove the extern "C", the function name exported would be: ?create@@YA?AV?$list@PAUanalyzer@@V?$allocator@PAUanalyzer@@@std@@@std@@XZ

I want the name to be clean, so I added extern "C", now it conflicts

Any other way can get a clean function name?

Thanks.

Upvotes: 2

Views: 843

Answers (2)

John Duffy
John Duffy

Reputation: 334

You can return a pointer instead of the object.

extern "C" __declspec(dllexport) std::list<std::string>* createAllocated()
{
    return new std::list<std::string>();
}

Due to the nature of C++, the caller will need to have the compatible ABIs.

Upvotes: 1

Some programmer dude
Some programmer dude

Reputation: 409136

When you say extern "C" you tell the compiler to create a function that can be called from other languages than C++ (most notably C). However, no other language than C++ have std::list, so it can not create such a function because the return type is part of the function signature in C++. And if you don't create a function which has C-compatible return type (or arguments) you can not create an extern "C" function.

If you're going to use the DLL from a C++ program there is no need for the extern "C" part. The C++ compiler and linker will be able to handle the mangled name without problems anyway.

Upvotes: 3

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