Why in c++, a derived class template publicly Inherited from base class template can't call the base class' funtion directly?

Asked Mar 06 '20 at 08:04

Active Mar 06 '20 at 08:08

Viewed 36 times

This is the code:

class A : public std::vector<int>
{
  public:
    void fooA()
    {
        std::cout << "A::fooA() + " << size() << '\n';
    }
};

template <typename T>
class B : std::vector<T>
{
  public:
    void fooB()
    {
        std::cout << "B::fooB() + " << size() << '\n';
    }
};

Class A is correct. But in class B, size() is considered undeclared, unless I change it to std::vector<T>::size()

Why doesn't class A have to add std::vector<int>, but does class B have to?

asked Mar 06 '20 at 08:04

xxhxx

1

Alternatively, you can write `this->size()`. – JFMR Mar 06 '20 at 08:05
Yes, it is more simple, but still can't call it directly. – xxhxx Mar 06 '20 at 08:10
FYI, on an unrelated note, it is not a good idea to derive from standard containers. They are not designed for inheritance. For instance, they don't have virtual destructors. – Remy Lebeau Mar 06 '20 at 08:10

Why in c++, a derived class template publicly Inherited from base class template can't call the base class' funtion directly?

0 Answers0