protected inheritance - why a certain command is not allowed

Question

Suppose I have the following code:

class A
    {
    public:
        void foo() const {}
    };

    class B : protected A
    {
    public:
    void print() const
    {
        foo();
    }
    };

    void main()
    {

    B b;
    b.print();
    b.foo();
    }

Now, by reading Difference between private, public, and protected inheritance, I conclude that in case of protected inheritance, every public member of the base (for that matter - class A) will be acsessible in the derived class (class B).

However, I dont understand why the command b.foo(); is not allowed in this case, becuase it apparently seems to be allowed according to the rules of protected inheritance.

It might help to pretend `B` declares a `protected: void foo() const;`. The same access restrictions apply. — François Andrieux, Feb 22 '19 at 18:44
The access rules state that `B` can access `foo` but in `b.foo();` it's not `B` doing the accessing but you — NathanOliver, Feb 22 '19 at 18:44
**in** is the keyword. The scope of main is not in there, is it? — StoryTeller - Unslander Monica, Feb 22 '19 at 18:44

score 0 · Answer 1 · answered Feb 22 '19 at 18:58

0

You are trying to access B::foo() from the scope of main(). As foo is not public in this context, it's not allowed.

answered Feb 22 '19 at 18:58

pablo285

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score -1 · Answer 2 · answered Feb 22 '19 at 19:07

-1

In this case, only the class B "knows" or "is aware" of the inherent relationship between class A and itself due to the relationship being protected. Namely, main() doesn't know of said relationship.

answered Feb 22 '19 at 19:07

Anzurio

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protected inheritance - why a certain command is not allowed

2 Answers2