CODEWITHSUNDEEP
c++c++11templatestemplate-specialization
Daniel Saad
Daniel Saad

Reputation: 425

Specializing some methods in template

I have a class which has a template specialization version. However, the former can't see the method implemented by the generic version. How can I make all methods in the generic version visible by the specialized version?

For example:

test.hpp

#include <iostream>

template <typename T>
class A_base{
public:
    virtual void foo() = 0;
};

template <typename T>
class A : public A_base<T> {
public:
    void foo() override {
        std::cout << "foo: generic type" << "\n";
    }
};

template <>
class A<int> : public A_base<int>{
public:
    void bar() {
        std::cout << "bar: int type" << "\n";
    }
};

test.cpp

#include "test.hpp"

int main(){
    A<int> a;
    a.foo(); // expected "foo: generic type"
    a.bar(); // expected "bar: int type"
}

Why A<int> a can't see foo()?

Upvotes: 2

Views: 45

Answers (1)

jfMR
jfMR

Reputation: 24738

Why A<int> a can't see foo()?

By specializing the class template A<T> for T = int, you are defining how the class template A<T> is when T corresponds to int, and that specialization (i.e.: A<int>) you are providing has no member called foo (the primary template however does).

It is possible to individually specialize the member functions of a class template. So, you can simply specialize the member function bar of the class template T for T = int instead of doing it for the whole class template:

template <>
void A<int>::bar(){
        std::cout << "bar: int type" << "\n";
}

Upvotes: 2

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