How to deduce types by template integers in C++

February 28, 2023

I have code:

template<int I, typename = void>
struct indexed_type {
  using type = void;
  indexed_type() { std::cout << "I is void: " << I << std::endl; }
};


template<int I>
struct indexed_type<I, typename std::enable_if<I == 1, char>::type> {
  using type = char;
  indexed_type() { std::cout << "I is char: " << I << std::endl; }
};

But after an initialization:

indexed_type<0>{};
indexed_type<1>{}; // not happens always

I have got not worked second variant always 🙁

Could you help me to get working the second variant of specialization ?

—
Thanks

>Solution :

Your specialization will only be used if the second type is given and is char, like indexed_type<1, char>. Without specifying the second parameter, it defaults to void (indexed_type<1> is the same as indexed_type<1, void>, which does not select the second specialization).

You probably want to specialize based on the value of I:

template<int I>
struct indexed_type;

template<>
struct indexed_type<0> {
  using type = void;
  static constexpr int I = 0;
  indexed_type() { std::cout << "I is void: " << I << std::endl; }
};

template<>
struct indexed_type<1> {
  using type = char;
  static constexpr int I = 1;
  indexed_type() { std::cout << "I is char: " << I << std::endl; }
};