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 void foo(T& t);  // A
void foo(Base&          


        

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 
#include 

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

void foo(Base& x)
{
}