Prevent implicit template instantiation

Question

In a method overload situation like this:

struct A
{
  void foo( int i ) { /*...*/ }
  template<typename T> void foo( T t ) { /*...*/ }
}

How can I prevent template instantiation unless explicitly commanded?:

A a;
a.foo<int>( 1 ); // ok
a.foo<double>( 1.0 ); // ok
a.foo( 1 ); // calls non-templated method
a.foo( 1.0 ); // error

Thanks!

Answer 1

You can introduce a depedent_type struct that prevents template argument deduction.

template <typename T>
struct dependent_type
{
    using type = T;
};

struct A
{
  void foo( int i ) { /*...*/ };
  template<typename T> void foo( typename dependent_type<T>::type t ) { /*...*/ }
}

Which in your example:

a.foo<int>( 1 );      // calls the template
a.foo<double>( 1.0 ); // calls the template
a.foo( 1 );           // calls non-templated method
a.foo( 1.0 );         // calls non-templated method (implicit conversion)

wandbox example

(This behavior is explained on cppreference > template argument deduction > non-deduced contexts.)

---

If you want to make a.foo( 1.0 ) a compilation error, you need to constrain the first overload:

template <typename T> 
auto foo( T ) -> std::enable_if_t<std::is_same<T, int>{}> { }

This technique makes the above overload of foo take only int arguments: implicit conversions (e.g. float to int) are not allowed. If this is not what you want, consider TemplateRex's answer.

wandbox example

(With the above constrained function, there is a curious interaction between the two overloads when a.foo<int>( 1 ) is called. I asked a question about it as I'm not sure of the underlying rules that guide it.)