Why is not overloaded function for derived class object invoked when given a pointer to base class in C++?

Question

In the following code

#include <iostream>
using namespace std;

class A {
  public:
    A() {}
    virtual ~A() {};
};

class B : public A {
  public:
    B() {}
    virtual ~B() {};
};

void process(const A&) {
    cout << "processing A" << endl;
}

void process(const B&) {
    cout << "processing B" << endl;
}

int main(void) {
    A* a = new B;
    process(*a);
    return 0;
}

the output of running it becomes

processing A

but I would have assumed that it should have been

processing B

since a points to the derived class B and not A. So why does it call the first implementation of process function and not the second?

Vlad from Moscow · Accepted Answer · 2014-03-26 12:32:44Z

2

The static type of expression *a is A because a was declared as

A* a = new B;

The compiler resolves the selection of overloaded functions using the static type of the argument.

Even when virtual functions are called the compiler uses the static type of the object to call appropriate function. The difference is only that the compiler uses the table of pointers to virtual functions to indirectly call the required function.

edited Mar 26, 2014 at 12:32

answered Mar 26, 2014 at 10:42

Vlad from Moscow

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7 Comments

iavr Over a year ago

The type of *a is actually A&.

Vlad from Moscow Over a year ago

@iavr From the C++ Standard: "If the type of the expression is “pointer to T,” the type of the result is “T.”

iavr Over a year ago

Instantiating decltype(*a)::dummy() results in 'decltype(*a)' (aka 'A &') is not a class, namespace, or scoped enumeration. Same for decltype((*a)). Maybe I'm missing something?

Vlad from Moscow Over a year ago

@iavr You missed the quote of the Standard I cited.

Vlad from Moscow Over a year ago

@iavr The compiler returns lvalue that is the same object as the original that is refered to. This reference has type T.

|

iavr · Accepted Answer · 2014-03-26 11:07:43Z

2

You need to make process() a virtual member function of A, B:

class A {
  public:
    A() {}
    virtual ~A() {};
    virtual void process() const { cout << "processing A" << endl; }
};

class B : public A {
  public:
    B() {}
    virtual ~B() {};
    virtual void process() const override { cout << "processing B" << endl; }
};

int main(void) {
    A* a = new B;
    a->process();
    return 0;
}

In your current code, *a is of type A&, so the closest match to process(*a); is the first overload (for const A&).

edited Mar 26, 2014 at 11:07

answered Mar 26, 2014 at 10:41

iavr

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Comments

jrok · Accepted Answer · 2014-03-26 10:41:21Z

1

void process(const A&); is a better (exact) match, since dereferencing A* gives you A&.

Short answer, but there isn't much more to say unless you want a reference from the standard.

You could dynamic_cast the result of *a and that would give you a B&, but that's smelly desing. What you probably want is a virtual function in A that's overriden in B (assume it's called foo). Then, calling a->foo() would dispatch to B::foo.

answered Mar 26, 2014 at 10:41

jrok

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