具有可变模板的给定成员函数指针的类型推导

你至少需要一个助手和类似的东西 std::tuple :

template<typename R, typename C, typename... A>
std::tuple<A...> deduce_A_tuple(R(C::*)(A...));

template<typename R, typename C, typename T>
struct FooHelper;

template<typename R, typename C, typename... A>
struct FooHelper< R, C, std::tuple<A...> >
{
    typedef Foo< R( C, A... ) > type;
};

然后您可以使用:

FooHelper< decltype(deduce_R_type(&SomeClass::SomeFunction)),
           decltype(deduce_C_type(&SomeClass::SomeFunction)), 
           decltype(deduce_A_tuple(&SomeClass::SomeFunction)) >::type instance;

Live example

作为用户 DyP 指出,当你无论如何都必须使用助手时,这可能会简单得多:

template<typename>
struct FooHelper;

template<typename R, typename C, typename... A>
struct FooHelper< R (C::*)(A...) >
{
    using type = Foo< R( C, A... ) >;
};

并将其用作

FooHelper< decltype(&SomeClass::SomeFunction) >::type instance;