CODEWITHSUNDEEP
c++
Ayush
Ayush

Reputation: 42450

Simple inheritance in C++

I was going over some sample questions for an upcoming test, and this question is totally confusing me.

Consider the following code:

class GraduateStudent : public Student 
{
  ...
};

If the word "public" is omitted, GraduateStudent uses private inheritance, which means which of the following?

  1. GraduateStudent objects may not use methods of Student.

  2. GraduateStudent does not have access to private objects of Student.

  3. No method of GraduateStudent may call a method of Student.

  4. Only const methods of GraduateStudent can call methods of Student.

Upvotes: 7

Views: 531

Answers (4)

Matthieu M.
Matthieu M.

Reputation: 299800

First: the inheritance access specifier does not modify the way Derived interacts with Base, it modifies the way the world can treat a Derived object as if it were a Base.

Consider:

struct A{};
struct B: xxxx A
{
  friend void friendly();
};
struct C: B {};

void outsider();

A very simple table: I summarize which zones of A can be accessed by the various actors.

xxxx         public      protected      private

B            pub/prot    pub/prot       pub/prot

friendly     pub/prot    pub/prot       pub/prot    *

C            pub/prot    pub/prot       --

outsider     pub         --             --

Note (*): A friend has the same rights that the object itself, no surprise there.

A simple rule of thumb is to consider that the inheritance access specifier overrides the access specifiers of the base class if they are looser.

Since nothing is looser than public, it does not change anything. protected means that the public section of the Base becomes protected, and private means that the public and protected sections become private.

Upvotes: 1

Daniel Trebbien
Daniel Trebbien

Reputation: 39208

The only correct answer that I see is 2, but that does not depend on whether GraduateStudent inherits from Student privately or publicly.

Private inheritance means that it is private knowledge that GraduateStudent is a Student, so only friends of GraduateStudent and GraduateStudent members can static_cast GraduateStudent& to Student& and GraduateStudent* to Student*, access public or protected member variables of Student, and call public or protected member functions of Student.

See: http://www.parashift.com/c++-faq-lite/private-inheritance.html

Upvotes: 3

Tyler McHenry
Tyler McHenry

Reputation: 76660

Although this is a bare homework-ish question, I'm going to answer it because it is a terrible question. I would almost consider it a trick question, and it doesn't really make for a good test of knowledge.

The answer is 2. GraduateStudent does not have access to private objects of Student., except that this has nothing at all to do with private inheritance. Point 2 would be true whether or not the public keyword were present, since derived classes never have access to the private members of their base classes, no matter how they inherit.

Private inheritance means essentially two things (as opposed to public inheritance):

  1. All public methods of Student become private methods in GraduateStudent. That means that if, for example, Student has a public method foo(), then GraduateStudent has a private method foo().

  2. The base class is "inaccessible", which means that polymorphism does not work. In layman's terms, this means that if GraduateStudent inherits privately from Student, then you cannot treat a GraduateStudent* as if it were a Student* (or a GraduateStudent& as if it were a Student&).

It's possible that the author of the question also meant for point 1 to be a correct answer, but it is ambiguously worded. What does it mean that "GraduateStudent objects may not use methods of Student"? It's possible that the intent is for this to mean that you cannot call methods inherited from Student on objects of type GraduateStudent, like I wrote in the first point above, but the GraduateStudent object itself, within its methods, can use methods of Student.

For example:

class Student {
  public: 
    void foo() {};
};

class GraduateStudent : Student {
  public: 
   void bar()
   {
     foo(); // Legal
   }
};

int main() {
  GraduateStudent g;

  g.bar(); // Legal
  g.foo(); // Illegal

  return 0;
};

Upvotes: 22

display101
display101

Reputation: 2085

1. GraduateStudent objects may not use methods of Student.

Not true, all GraduateStudent objects can use any public or protected members of Students (obviously private members are the exception here) internally. Also any outsider using these objects cannot access the base Student class of the object, the access to the base class must happen within the context of the GraduateStudent methods.

2. GraduateStudent does not have access to private objects of Student.

Yes

3. No method of GraduateStudent may call a method of Student.

Not true

4. Only const methods of GraduateStudent can call methods of Student.

Nope, there's no distinction made on const members having more access to the base class than non-const members would have.

Upvotes: 3

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