在以下(有效的)代码示例中，模板化的 register_enum() 函数用于迭代枚举并调用用户提供的回调以将枚举值转换为 C 字符串。所有枚举都在一个类中定义，枚举到字符串的转换是使用静态 to_cstring( enum ) 函数完成的。当一个类(如下面的着色器类)有多个枚举和相应的重载 to_cstring(enum) 函数时，编译器无法决定将哪个是正确的 to_cstring() 函数传递给 register_enum()。我认为代码比我能解释得更好...

#include <functional>
#include <iostream>

// Actual code uses Lua, but for simplification
// I'll hide it in this example.
typedef void lua_State;

class widget
{
    public:
        enum class TYPE
        {
            BEGIN = 0,
            WINDOW = BEGIN,
            BUTTON,
            SCROLL,
            PROGRESS,
            END
        };

        static const char *to_cstring( const TYPE value )
        {
            switch ( value )
            {
                case TYPE::WINDOW: return "window";
                case TYPE::BUTTON: return "button";
                case TYPE::SCROLL: return "scroll";
                case TYPE::PROGRESS: return "progress";
                default: break;
            }
            return nullptr;
        }
};

class shader
{
    public:
        enum class FUNC
        {
            BEGIN = 0,
            TRANSLATE = BEGIN,
            ROTATE,
            SCALE,
            COLOR,
            COORD,
            END
        };

        enum class WAVEFORM
        {
            BEGIN = 0,
            SINE = BEGIN,
            SQUARE,
            TRIANGLE,
            LINEAR,
            NOISE,
            END
        };

        static const char *to_cstring( const FUNC value )
        {
            switch ( value )
            {
                case FUNC::TRANSLATE: return "translate";
                case FUNC::ROTATE: return "rotate";
                case FUNC::SCALE: return "scale";
                case FUNC::COLOR: return "color";
                case FUNC::COORD: return "coord";
                default: break;
            }
            return nullptr;
        }

        static const char *to_cstring( const WAVEFORM value )
        {
            switch ( value )
            {
                case WAVEFORM::SINE: return "sine";
                case WAVEFORM::SQUARE: return "square";
                case WAVEFORM::TRIANGLE: return "triangle";
                case WAVEFORM::LINEAR: return "linear";
                case WAVEFORM::NOISE: return "noise";
                default: break;
            }
            return nullptr;
        }
};

// Increment an enum value.
// My compiler (g++ 4.6) doesn't support type_traits for enumerations, so
// here I just static_cast the enum value to int... Something to be fixed
// later...
template < class E >
E &enum_increment( E &value )
{
    return value = ( value == E::END ) ? E::BEGIN : E( static_cast<int>( value ) + 1 );
}

widget::TYPE &operator++( widget::TYPE &e )
{
    return enum_increment< widget::TYPE >( e );
}

shader::FUNC &operator++( shader::FUNC &e )
{
    return enum_increment< shader::FUNC >( e );
}

shader::WAVEFORM &operator++( shader::WAVEFORM &e )
{
    return enum_increment< shader::WAVEFORM >( e );
}


// Register the enumeration with Lua
template< class E >
void register_enum( lua_State *L, const char *table_name, std::function< const char*( E ) > to_cstring )
{
    (void)L; // Not used in this example.
    // Actual code creates a table in Lua and sets table[ to_cstring( i ) ] = i
    for ( auto i = E::BEGIN; i < E::END; ++i )
    {
        // For now, assume to_cstring can't return nullptr...
        const char *key = to_cstring( i );
        const int value = static_cast<int>(i);
        std::cout << table_name << "." << key << " = " << value << std::endl;
    }
}

int main( int argc, char **argv )
{
    (void)argc; (void)argv;

    lua_State *L = nullptr;

    // Only one to_cstring function in widget class so this works...
    register_enum< widget::TYPE >( L, "widgets", widget::to_cstring );

    // ... but these don't know which to_cstring to use.
    register_enum< shader::FUNC >( L, "functions", shader::to_cstring );
    //register_enum< shader::WAVEFORM >( L, "waveforms", shader::to_cstring );

    return 0;
}

编译器输出:

$ g++ -std=c++0x -Wall -Wextra -pedantic test.cpp -o test && ./test
test.cpp: In function ‘int main(int, char**)’:
test.cpp:140:69: error: no matching function for call to ‘register_enum(lua_State*&, const char [10], <unresolved overloaded function type>)’
test.cpp:140:69: note: candidate is:
test.cpp:117:7: note: template<class E> void register_enum(lua_State*, const char*, std::function<const char*(E)>)

如何将正确的 to_cstring 函数传递给 register_enum()？我意识到我可以重命名各个 to_cstring() 函数，但我想尽可能避免这种情况。也许我的设计很臭，你可以推荐一个更好的方法。

我的问题类似于 Calling overloaded function using templates (unresolved overloaded function type compiler error)和 How to get the address of an overloaded member function?但到目前为止，我无法将该信息应用于我的具体问题。

最佳答案

错误告诉您有两个潜在的重载可以使用，编译器无法为您决定。另一方面，您可以通过强制转换来确定使用哪个:

typedef const char *(*func_ptr)( shader::FUNC );
register_enum< shader::FUNC >( L, "functions", (func_ptr)shader::to_cstring );

或者没有 typedef(在更难阅读的单行代码中):

register_enum< shader::FUNC >( L, "functions", 
             (const char *(*)( shader::FUNC ))shader::to_cstring );

^*请注意，在函数签名中，顶层 const被移除。

下一个问题是为什么编译器自己没有找到合适的重载？问题是在对 register_enum 的调用中您传递枚举的类型，这决定了 std::function 的类型成为std::function< const char* ( shader::FUNC ) > ，但是std::function有一个模板化的构造函数，在尝试推断构造函数的参数类型之前，编译器必须知道您要使用哪个重载。

关于c++ - 使用 std::function 时如何解决此 <unresolved overloaded function type> 错误？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10038259/