创建N个零的索引序列

我一直想写一本书 zero_sequence 创建 index_sequence 以产生一个新的具有相同数量的值,但都是零。我很困惑为什么会这样:

template<typename> struct zero_sequence;
template<size_t...I> 
struct zero_sequence<index_sequence<I...>> : index_sequence<(I*0u)...>{};

int main( int argc, char *argv[] ) {

  using A  = index_sequence<0,1,2,3,4>;
  using B  = make_index_sequence<5>;
  using A0 = zero_sequence<A>;
  using B0 = zero_sequence<B>;

  A a;  B b; 

  cout <<< std::is_same<A,B>::value << endl; // is false

  A0 a0; 
  B0 b0;  // < error implicit instantiation of undefined template
  return 0;
}

我不明白为什么这两种情况是不同的(在macclangllvm9.0.0上)。

如果它与解释这种行为有关,我将使用 和 make_index_sequence 在C++ 11中 I think I got it from here

template<typename Integer, Integer... I>
struct integer_sequence {
  using type = integer_sequence;
  using value_type = Integer;
  static constexpr size_t size() noexcept { return sizeof...(I); }
};

template<size_t... Ints> using index_sequence = integer_sequence<size_t, Ints...>;

template <typename,typename> struct _merge_and_renumber;

template <size_t... I1, size_t... I2>
struct _merge_and_renumber<index_sequence<I1...>,index_sequence<I2...>> 
  : index_sequence<I1..., (sizeof...(I1)+I2)...> {};

template <size_t N>
struct make_index_sequence
  : _merge_and_renumber<typename make_index_sequence<N/2>::type,
                        typename make_index_sequence<N - N/2>::type> {};

template<> struct make_index_sequence<0> : index_sequence<> {};
template<> struct make_index_sequence<1> : index_sequence<0> {};