Access members of a class within a different class in c++

Question

I am trying to understand object oriented programming using c++. The following is a minimal example for which the result is not what I naively expect:

#include <iostream>

class B {
public:
      B (int val) : val(val) {;}

      int get_val() { return val; }
      int set_val(int a) { val = a; }

private:
      int val;
};

class A {
public:
      A (B b) : b(b) {;}
      B get_b() { return b; }

private:
      B b;
};

int main(){
    B b_main(5);
    std::cout << b_main.get_val() << std::endl; // Prints 5, which makes sense

    A a_main(b_main);
    std::cout << a_main.get_b().get_val() << std::endl; // Prints 5, which makes sense

    a_main.get_b().set_val(2);
    std::cout << a_main.get_b().get_val() << std::endl; // Why does this not print 2?

    return 0;
}

The last cout statement does not make sense to me. In the second to last line, I set the value of the object to be 2, so why does this not print 2? Looking at some similar questions on Stack Exchange, I found some suggestions to make A and B be friends of each other. I tried adding friend class A in class B and friend class B in class A, but this did not work. In my understanding, adding the friend statements should be unnecessary since I have the get_b() method in class A. I found some suggestions to try passing the object of type B in by reference to the constructor of A: A (B& b) : b(b) {;} but this did not work either.

Can anyone explain to me why the program is not producing the intended result and also how to obtain the desired result (that is, the last cout statement prints 2)?

Note: I also experimented with the following. I made the private variable b of class A be public:

#include <iostream>

class B {
public:
    B (int val) : val(val) {;}
    int get_val() { return val; }
    int set_val(int a) { val = a; }

private:
    int val;
};

class A {
public:
  A (B b) : b(b) {;}
  B b; // This is now public
  //B get_b() { return b; } // This is no longer needed

private:
};


int main(){
    B bmain(5);
    std::cout << bmain.get_val() << std::endl;

    A amain(bmain);
    std::cout << amain.b.get_val() << std::endl;

    amain.b.set_val(2);
    std::cout << amain.b.get_val() << std::endl; // Works!

    return 0;
}

And now I obtain the desired result. Is this how the code should be implemented as opposed to the first code snippet? I would like to have a get_b() method as in the first code snippet, but if this is not the correct way of going about this, please let me know.

score 3 · Accepted Answer · edited May 23 '17 at 11:45

In the second to last line, I set the value of the object to be 2, so why does this not print 2?

Because you return a copy of the B object in a_main with the get_b() method. What happens is that the b variable in a_main is copied, i.e. another object of class B, identical to the b member, is created, and returned to the caller. Then, that new B object is modified. But it has no connection to the original b in a_main. This has little to do with visibility and member access.

However, in the second example, you expose the b member in a_main and directly operate on that object without making a copy of it, thus the successful result. What the public modifier changes is that it allows you to access the b object directly, hence the effect.

I found some suggestions to try passing the object of type B in by reference to the constructor of A: A (B& b) : b(b) {;} but this did not work either.

That isn't going to work. What happens when you do so, is that the A::b is initialized using the value that is passed by reference, true. But the reference only leads to no additional copy of b passed to the constructor being made. This reference does not create a link between the b passed to the constructor and A::b. It's on the other end, so to say.

_{By the way, A (B& b) : b(b) {;} that the c'tor parameter name is identical to the member name is a bad practice. It's a good idea to have them named similarly, but still, add e.g. an underscore: A (B& _b) : b(_b) {;}}

If you want to achieve the same result in the first snippet, return a reference to b like so:

  B& get_b() { return b; }

Still, this is undesirable, because you expose a private member of class A just to allow clients of A to modify a certain property of that member. Better provide a method in A to set the val property of A::b without giving full access to A::b.

Definitely see this: What's the difference between passing by reference vs. passing by value?

and maybe this: Java and C++ pass by value and pass by reference

because I have a feel that you're coming from Java and expect pass-by-reference in C++ by default.

score 1 · Answer 2 · answered Feb 21 '16 at 19:08

get_b returns a copy of your private variable b, not the actual variable. If you want to be able to access it, you need to return a reference to b so that the returned value can be manipulated. Your get_b definition should look like this:

B& get_b() { return b; }

if that is what you expect to do. However, this is not usually a desirable solution. If you are going to be actively changing the value of b you should write a set_b function to manipulate the variable. And if you are really working with the variable a lot, reading and writing values to it, you should keep it public for fast access.

score 0 · Answer 3 · answered Feb 21 '16 at 19:18

Just for the sake of completeness, you can solve this problem as a C programing problem rather than using all the fancy references in C++ programing. When you get b_main from a_main, the returned object does not occupy the same memory address.

#include <iostream>

class B {

public:

  B (int val) : val(val) {;}

  int get_val() { return val; }
  int set_val(int a) { val = a; }

private:

  int val;

};

class A {

public:

  A (B b) : b(b) {;}
  B get_b() { return b; }

private:

  B b;

};


int main(){

  B b_main(5);
  B* addrb = &b_main;
   std::cout << b_main.get_val() << std::endl; // Prints 5, which makes sense
    std::cout<<"Address of b_main: "<<addrb<<std::endl;
  A a_main(b_main);
  B bt = a_main.get_b();
  addrb = &(bt);
  std::cout << a_main.get_b().get_val() << std::endl; // Prints 5, which makes sense
    std::cout<<"Address of a_main.get_b(): "<<addrb<<std::endl;
  a_main.get_b().set_val(2);
  std::cout << a_main.get_b().get_val() << std::endl; // Why does this not print 2?

  return 0;

}

Notice the difference in address of the new cout statements. One way to fix this is to return pointers rather than b itself. i.e.

#include <iostream>

class B {

public:

  B (int val) : val(val) {;}

  int get_val() { return val; }
  int set_val(int a) { val = a; }

private:

  int val;

};

class A {

public:

  A (B b) : b(b) {;}
  B* get_b() { return &b; }

private:

  B b;

};


int main(){

  B b_main(5);
  //B* addrb = &b_main;
   std::cout << b_main.get_val() << std::endl; // Prints 5, which makes sense
    //std::cout<<"Address of b_main: "<<addrb<<std::endl;
  A a_main(b_main);
  //B bt = a_main.get_b();
  //addrb = &(bt);
  std::cout << a_main.get_b()->get_val() << std::endl; // Prints 5, which makes sense
    //std::cout<<"Address of a_main.get_b(): "<<addrb<<std::endl;
  a_main.get_b()->set_val(2);
  std::cout << a_main.get_b()->get_val() << std::endl; // Why does this not print 2?

  return 0;

}

Access members of a class within a different class in c++

3 Answers3