Constructor conditionally marked explicit

Question

Update: conditional explicit has made it into the C++20 draft. more on cppreference

The cppreference std::tuple constructor page has a bunch of C++1

Answer 1

One way that seems to work with most compilers is to add a dummy parameter to one of the functions, to make them slightly different.

// constructor is explicit if T is integral

struct S {
  template ::value>::type>
  S(T t) {}

  template ::value>::type,
            typename dummy = void>
  explicit S(T t) {}
};

int main()
{
   S  s1(7);

   S  s2("Hello");    
}

Compiles with MSVC 2015.