在cpp文件中考虑以下问题:

struct someStruct{
    public:
    extern "C"  __declspec(dllexport) int sumup();
} someStruct;

extern "C" __declspec(dllexport) int someStruct::sumup()
{
    return 0;
}


无法编译:error: expected unqualified-id before string constant
是否可以导出具有C链接的C ++成员方法?

最佳答案

首先,链接规范不适用于成员函数。通过[dcl.link] / 4:


  [...]链接规范应仅在命名空间范围(3.3)中出现。 [...]


但是在同一段中,甚至在标准中甚至有一个示例与您的问题有些相关:


  在确定类成员名称和类成员函数的函数类型的语言链接时,将忽略C语言链接。 [例:

extern "C" typedef void FUNC_c();
class C {
  void mf1(FUNC_c*);      // the name of the function mf1 and the member
                          // function’s type have C ++ language linkage; the
                          // parameter has type pointer to C function
  FUNC_c mf2;             // the name of the function mf2 and the member
                          // function’s type have C ++ language linkage
  static FUNC_c* q;       // the name of the data member q has C ++ language
                          // linkage and the data member’s type is pointer to
                          // C function
};

extern "C" {
  class X {
    void mf();            // the name of the function mf and the member
                          // function’s type have C ++ language linkage
    void mf2(void(*)());  // the name of the function mf2 has C ++ language
                          // linkage; the parameter has type pointer to
                          // C function
  }
};

  
  —结束示例]

10-08 13:27
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