I have a class which has a partial 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 partially 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()?
Aucun commentaire:
Enregistrer un commentaire