Why is the compiler not selecting my function-template overload in the following example?
- by Steve Guidi
Given the following function templates:
#include <vector>
#include <utility>
struct Base { };
struct Derived : Base { };
// #1
template <typename T1, typename T2>
void f(const T1& a, const T2& b)
{
};
// #2
template <typename T1, typename T2>
void f(const std::vector<std::pair<T1, T2> >& v, Base* p)
{
};
Why is it that the following code always invokes overload #1 instead of overload #2?
void main()
{
std::vector<std::pair<int, int> > v;
Derived derived;
f(100, 200); // clearly calls overload #1
f(v, &derived); // always calls overload #1
}
Given that the second parameter of f is a derived type of Base, I was hoping that the compiler would choose overload #2 as it is a better match than the generic type in overload #1.
Are there any techniques that I could use to rewrite these functions so that the user can write code as displayed in the main function (i.e., leveraging compiler-deduction of argument types)?
