Base class method being called instead of derived, even when passed by reference

Question

I have 3 classes: A, B, and AnotherClass. Where B is derived from A:

class A {
public:
    A(){}
    virtual void method() {//default action}
};

Then I have a derived class, B:

class B : public A {
public:
    B(){}
    void method() {//redefine action}
};

And AnotherClass:

class AnotherClass {
public:
    AnotherClass(A& a);
    A a;
    anotherMethod(){ a.method()}
};
AnotherClass :: AnotherClass(A& a) : a(a) //initialization

So, if I construct an object of AnotherClass with an object of B:

B b();
AnotherClass myObj(b);

Keep in mind, since B inherits from A, and AnotherClass accepts an object of A, I am able to successfully pass in a B object as the argument.

And I call:

myObj.anotherMethod();

I expect this to execute anotherMethod(), and when it does, I expect it to call the REDEFINED method() that belongs to B, but instead it calls the default method() defined in A

I was thinking my problem is because I specify the argument of AnotherClass as an object of class A. However, I do not want to change this argument to an object of class B because I also have classes C, D, and E, that also inherit directly from A. So I want to use the base class as the argument type so I am not limited to only being able to pass in a b object. But, I read some older posts on this site and most proposed solutions was to pass the derived object (b) by reference, which I am doing.

Can someone explain why this is happening?

*I was thinking my problem is because I specify the argument of AnotherClass as an object of class A* You are correct. You need to store a reference or pointer to `A` in the class. — NathanOliver, Jun 09 '16 at 17:46
I've edited it, but not this: `B b();` - is your code even compiling? — Ajay, Jun 09 '16 at 18:18

score 7 · Accepted Answer · answered Jun 09 '16 at 17:57

The argument to the constructor is fine. It is an object of type B passed by reference to the A subobject. The problem is that you copy the A subobject only (and not the derived object) into the member variable a. Then when you call a.method() there is only a base object there, so the base function is called.

The solution is to make the member variable be a pointer or a reference (and not a copy) of the argument. (Note that in either case, you will then need to make sure that the object you pass in, will live at least as long as the AnotherClass object.

kemis · Answer 2 · 2016-06-09T18:57:10.457

2

It will only work if it is a pointer A* a Look up on polymorphism. http://www.cplusplus.com/doc/tutorial/polymorphism/

if you wont to know more , read the article

edited Jun 09 '16 at 18:57

answered Jun 09 '16 at 17:48

kemis

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`A *a(b);` is a compilation error. `A &a(b);` will also work. – Martin Bonner supports Monica Jun 09 '16 at 17:58
Wrong and completely unrelated to the problem (which is slicing when he stores the reference to an instance of A in the AnotherClass constructor). – kfsone Jun 09 '16 at 17:58
A *a=new A(b); this is nicer :) sorry – kemis Jun 09 '16 at 17:58
a) It will *not* work "only" for pointers. b) what you have marked as working will not work. -1. – Barry Jun 09 '16 at 18:45
`new A(b)` will copy-construct an `A` object from the base of the `b` object. `a->method()` will call the base class function, not the derived class function. – Martin Bonner supports Monica Jun 09 '16 at 19:00

score 2 · Answer 3 · answered Jun 09 '16 at 17:55

2

You are slicing in your constructor to AnotherClass. Note you have a value of type A as a member. So the compiler is doing the right thing. You don't have a B at the point you call a.method().

answered Jun 09 '16 at 17:55

Edward Strange

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Base class method being called instead of derived, even when passed by reference

3 Answers3

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