Why is the compiler choosing this template function over an overloaded non-template function?

Question

Using VC++ 2010, given the following:

class Base { };
class Derived : public Base { };

template<class T> void foo(T& t);  // A
void foo(Base& base);              // B

Derived d;
foo(d);                            // calls A
foo(static_cast<Base&>(d));        // calls B

I would like "B" to be called above. I can achieve this with a cast to Base, but why is this necessary?

I want the template function to be called for all types not derived from Base (built-in types, etc.), but I want the non-template overload to be called for types derived from Base, without requiring the client to explicitly cast. I also tried making the overload a specialization of the template, but the same behavior occurs in that case. What is the idiomatic way to get what I'm looking for?

Answer 1

All things being equal, nontemplate functions are preferred over function templates. However, in your scenario, all things are not equal: (A) is an exact match with T = Derived, but (B) requires a derived-to-base conversion of the argument.

You can work around this for specific cases (like this one) by using SFINAE (substitution failure is not an error) to prevent (A) from being instantiated with a type that is derived from Base:

#include <type_traits>
#include <utility>

template <typename T>
typename std::enable_if<
    !std::is_base_of<Base, T>::value
>::type foo(T& x)
{
}

void foo(Base& x)
{
}