模板专门化适用于g++，但不适用于Visual C++

Question

我有一堆在g++下编译得很好的模板化代码，但现在当我尝试在windows下使用Visual C++ 2010构建时，我得到了一堆错误。

我有一组模板函数，用于从Lua代码中获取和设置C++对象中的值。例如，我有这样的模板：

//        Class       Return type Getter function
template <typename T, typename U, U (T::*Getter)() const>
int luaU_get(lua_State* L)
{
    T* obj = luaW_check<T>(L, 1); // Gets userdata from stack and checks if it's of type T
    luaU_push(L, (obj->*Getter)()); // Runs the getter function specified in the template, and pushes the
    return 1;
}

(完整文件可在here中找到)

实例化如下：

static luaL_reg TextArea_MT[] =
{
    //                             Class     Return type   Getter function
    { "GetCharacterSize", luaU_get<TextArea, unsigned int, &TextArea::GetCharacterSize> },
    { NULL, NULL }
};

该getter的签名如下：

unsigned int GetCharacterSize() const;

我收到了一堆这样的错误：

2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2440: 'specialization' : cannot convert from 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const' to 'unsigned int *(__thiscall ag::ui::TextArea::* const )(void) const'
2>          Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2973: 'luaU_get' : invalid template argument 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const'
2>          C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\extern\LuaWrapper\LuaWrapperUtil.hpp(147) : see declaration of 'luaU_get'
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2440: 'specialization' : cannot convert from 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const' to 'unsigned int *ag::ui::TextArea::* const '
2>          There is no context in which this conversion is possible
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2973: 'luaU_get' : invalid template argument 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const'
2>          C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\extern\LuaWrapper\LuaWrapperUtil.hpp(131) : see declaration of 'luaU_get'
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2440: 'specialization' : cannot convert from 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const' to 'unsigned int ag::ui::TextArea::* const '
2>          There is no context in which this conversion is possible
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2973: 'luaU_get' : invalid template argument 'unsigned int (__thiscall ag::ui::TextArea::* )(void) const'
2>          C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\extern\LuaWrapper\LuaWrapperUtil.hpp(123) : see declaration of 'luaU_get'
2>C:\Users\Alex\Documents\Visual Studio 2010\Projects\game\dev\src\game\lua\LuaTextArea.cpp(103): error C2440: 'initializing' : cannot convert from 'overloaded-function' to 'lua_CFunction'
2>          None of the functions with this name in scope match the target type

Answer 1

这是VC++中的编译器错误。以下代码有效：

#include <iostream>

struct TextArea
{
    unsigned GetCharacterSize() const { return 0; }
};

template<typename T, typename U, U (T::*)() const>
int foo()
{
    return 1;
}

template<typename T, typename U, U* (T::*)() const>
int foo()
{
    return 2;
}

int main()
{
    std::cout << foo<TextArea, unsigned, &TextArea::GetCharacterSize>() << '\n';
}

并使用GCC 4.3.4、GCC 4.5.1和Comeau 4.3.10.1 Beta2 (无链接)进行编译，但在使用VC++ 2010 SP1时生成以下错误：

错误C2668：'foo'：对重载函数的调用不明确

EDIT:作为一种变通方法，它很难看，但我能想到的唯一一件事就是使用额外的间接层，这样就不会涉及重载：

#include <iostream>

struct WithPointer
{
    unsigned* GetCharacterSize() const { return nullptr; }
};

struct WithoutPointer
{
    unsigned GetCharacterSize() const { return 0u; }
};

template<bool UsePointerImplB>
struct kludge
{
    template<typename T, typename U, U (T::*Getter)() const>
    static int foo() { return 1; }
};

template<>
struct kludge<true>
{
    template<typename T, typename U, U* (T::*Getter)() const>
    static int foo() { return 2; }
};

int main()
{
    std::cout
        << kludge<false>::foo<WithoutPointer, unsigned, &WithoutPointer::GetCharacterSize>() << '\n'
        << kludge<true>::foo<WithPointer, unsigned, &WithPointer::GetCharacterSize>() << '\n';
}

实际上，这与给每个重载一个不同的名称没有什么不同。

Answer 2

如果您可以强制用户选择函数的实际返回类型，则可以使用以下方法。也许，它会对你有用：

#include <iostream>

struct FooBar
{
  int Foo( void ) const
  {
   std::cout << "FooBar::Foo()" << std::endl;
   return ( 0 );
  }
  int * Bar( void ) const
  {
   std::cout << "FooBar::Bar()" << std::endl;
   return ( 0 );
  }
};

template< typename P00, typename P01, P01(P00::*p02)( void ) const >
void Call()
{
 P00 lT;
 ( lT.*p02 )();
}

int main( void )
{
 Call< FooBar, int, &FooBar::Foo > ();
 Call< FooBar, int*, &FooBar::Bar > ();

 return( 0 );
}

程序输出：

FooBar::Foo()
FooBar::Bar()