Question

Why does compiler allow this?

#include <iostream>
using namespace std;

template<typename T>
void func(T t)
{
    cout<<"The value is :"<<t<<"\n";
}

template<>
void func<int>(int d)    //Template Specialization
{
    cout<<"Template function\n";
}

void func(int d)        //Non-template function with same name & signature
{
    cout<<"Non-template function\n";
}

int main()
{
    func(4);
    func(4.67);
    func("TENE");
}

Answer 1

The answer to the question lies in section 13.3.3

Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then

— for some argument j, ICSj(F1) is a better conversion sequence than ICSj(F2), or, if not that,

— F1 is a non-template function and F2 is a function template specialization, or, if not that,

— F1 and F2 are function template specializations, and the function template for F1 is more specialized than the template for F2 according to the partial ordering rules described in 14.5.5.2, or, if not that, …

So amongst the three choices (non-templated func, specialization of func<int>, and the generic template func<typename T>, the non-template version of func is the best. Since there is a perfect match here, there is no reason for the compiler to even look to the templated versions of the function.