Question

In template meta programming, one can use SFINAE on the return type to choose a certain template member function, i.e.

template<int N> struct A {
  int sum() const noexcept
  { return _sum<N-1>(); }
private:
  int _data[N];
  template<int I> typename std::enable_if< I,int>::type _sum() const noexcept
  { return _sum<I-1>() + _data[I]; }
  template<int I> typename std::enable_if<!I,int>::type _sum() const noexcept
  { return _data[I]; }
};

However, this doesn’t work on constructors. Suppose, I want to declare the constructor

template<int N> struct A {
   /* ... */
   template<int otherN>
   explicit(A<otherN> const&); // only sensible if otherN >= N
};

but disallow it for otherN < N.

So, can SFINAE be used here? I’m only interested in solutions which allow automatic template-parameter deduction, so that

A<4> a4{};
A<5> a5{};
A<6> a6{a4};  // doesn't compile
A<3> a3{a5};  // compiles and automatically finds the correct constructor

Note: this is a very simplified example where SFINAE may be overkill and static_assert may suffice. However, I want to know whether I can use SFINAE instead.

Answer 1

You can add a defaulted type argument to the template:

template <int otherN, typename = typename std::enable_if<otherN >= N>::type>
explicit A(A<otherN> const &);