我正在写一个 dll，它导出一个函数
extern "C"
__declspec(dllexport)
std::list<string> create() {
return std::list<string>();
}
编译器抱怨：
错误 C2526：“创建”：C 链接函数无法返回 C++ 类“std::list<_Ty>”
如果我删除外部“C”，导出的函数名称将是：?create@@YA?AV?$list@PAUanalyzer@@V?$allocator@PAUanalyzer@@@std@@@std@@XZ
我希望名称干净，所以我添加了 extern "C"，现在它冲突了
还有其他方法可以获得干净的函数名称吗？
谢谢。
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.
您可以返回指针而不是对象。
extern "C" __declspec(dllexport) std::list<std::string>* createAllocated()
{
return new std::list<std::string>();
}
由于 C++ 的性质，调用者需要具有兼容的 ABI。