C++ function template full specialization error

Question

I am having this template matching error. I know partial specialization is not allowed in for function template, but I think it should work with full specialization. What change do I need to make to fix this issue ? Thanks.

#include <iostream>

template<typename T>
void allocate(){
    std::cout << "default" << std::endl;
}

template<>
void allocate<int>() {
    std::cout << "int" << std::endl;
}

template<>
void allocate<double>() {
    std::cout << "double" << std::endl;    
}




int main()
{
    allocate();  // Compiler error, I expect this should match the first template function.
    allocate<int>();
    allocate<double>();
    
    return 0;
}

Answer 1

You need to specify the template argument explicitly, the template parameter T can't be deduced from the context. eg

allocate<void>();

Or specify default argument for the template parameter, eg

template<typename T = void>
void allocate(){
    std::cout << "default" << std::endl;
}

then you can call it as

allocate(); // T is void

Answer 2

The primary template needs the template parameter to be specified explicitly, so you can do:

allocate<struct T>();  // ok

and since T is a new type named only for the purpose of this call, there is guaranteed to not be a specialization for this type, and the primary template will be called.

---

You could also give a default type for the template parameter in the primary:

template<typename T = struct Default>
void allocate(){
    std::cout << "default" << std::endl;
}

and again, no specialization can exist for Default since this type only exists in the scope of the primary template.

Now you can call the primary template without template parameters:

allocate();