c++ - 为什么显式给出的模板参数不能为 “deduced”

Question

来自这个问题:
Using enum values in combination with SFINAE
我尝试实现:

enum Specifier
{
    One,
    Two,
    Three
};

template <Specifier, typename UNUSED=void>
struct Foo 
{
        void Bar(){ std::cout << "Bar default" << std::endl;}
};

template <Specifier s , typename std::enable_if<s == Specifier::Two || s == Specifier::One, int>::type>
struct Foo<s>
{
    void Bar(){ std::cout << "Bar Two" << std::endl; }
};


int main()
{
   Foo< One >().Bar();
   Foo< Two >().Bar();
}

失败:

> main.cpp:130:8: error: template parameters not deducible in partial specialization:
  130 | struct Foo<s>
      |        ^~~~~~
   main.cpp:130:8: note:         '<anonymous>'

如何解决这个 super 简单的例子？我喜欢SFINAE :-)

Answer 1

将enable_if放入Foo的模板参数列表中:

template <Specifier s>
struct Foo<s, typename std::enable_if<s == Specifier::Two || s == Specifier::One, void>::type>
//                                           same as the default type used before ^^^^

demo。