What exactly does runtime polymorphism mean?

Question

I'm slightly confused about runtime polymorphism. Correct me if I am wrong, but to my knowledge, runtime polymorphism means that function definitions will get resolved at runtime.

Take this example:

class a
{
a();
~a();
void baseclass();
}

class b: class a
{
b();
~b();
void derivedclass1();
}

class c: class a
{
c();
~c();
void derivedclass2();
}

Calling methodology:

b derived1;
a *baseptr = &derived1; //here base pointer knows that i'm pointing to derived class b. 
baseptr->derivedclass1();

In the above calling methodology, the base class knows that it's pointing to derived class b.

So where does the ambiguity exist?

In what cases will the function definitions get resolved at runtime?

What ambiguity? The code you posted does not exhibit polymorphism of any kind. — , Aug 11 '10 at 09:49
As a matter of fact the code does not compile because you can't call `derivedclass1` using a pointer to `a` (unless you explicitly cast to `b` first, of course). — sepp2k, Aug 11 '10 at 09:50

score 10 · Answer 1 · 2010-08-11T09:56:08.263

This code, at run time, calls the correct version of f() depending on the type of object (A or B) that was actually created - no "ambiguity". The type cannot be known at compile-time, because it is selected randomly at run-time.

struct A {
   virtual ~A() {}
   virtual void f() {}
};

struct B : public A {
   virtual void f() {}
};


int main() {
   A * a = 0;
   if ( rand() % 2 ) {
      a = new A;
   }
   else {
      a = new B;
   } 
   a->f();   // calls correct f()
   delete a;
}

score 5 · Accepted Answer · answered Aug 11 '10 at 09:56

There is no ambiguity exists in the example provided.

If the base class has the same function name as the derived class, and if you call in the way you specified, it will call the base class's function instead of the derived class one.

In such cases, you can use the virtual keyword, to ensure that the function gets called from the object that it is currently being pointed. It is resolved during the run time.

Here you can find more explanation..

your link seems does not work. pls correct it. Thanks. – uss Nov 17 '15 at 13:06 — uss, Nov 17 '15 at 13:06

DumbCoder · Answer 3 · 2010-08-11T09:58:19.760

1

Turn this

void baseclass();

to

virtual void baseclass();

Override this in your Derived classes b and c. Then

b *derived1 = new derived1 ();
a *baseptr = derived1; //base pointer pointing to derived class b. 
baseptr->baseclass();

will invoke derived1 definition, expressing run time polymorphism. And do remember about making your destructor virtual in Base. Some basic reading material for polymorphism

edited Aug 11 '10 at 09:58

answered Aug 11 '10 at 09:52

DumbCoder

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cybevnm · Answer 4 · 2010-08-11T10:09:26.250

Runtime means that exact method will be known only at run time. Consider this example:


class BaseClass
{
public:
  virtual void method() {...};
};

class DerivedClassA : public BaseClass
{
  virtual void method() {...};
};

class DerivedClassB : public BaseClass
{
  virtual void method() {...};
};

void func(BaseClass* a)
{
  a->method();
}

When you implement your ::func() you don't know exactly type of instance pointed by BaseClass* a. It might be DerivedClassA or DerivedClassB instance etc.
You should realize, that runtime polymorphism requires special support from language (and maybe some overhead for calling "virtual" functions). In C++ you "request" for dynamic polymorphism by declaring methods of base class "virtual" and using public inheritance.

Tunvir Rahman Tusher · Answer 5 · 2013-05-29T12:03:45.380

Lets have an experiment

#include <iostream>
using namespace std;
class aBaseClass
{

public:

    void testFunction(){cout<<"hello base";}///Not declared as virtual!!!!

};
class aDerivedClass:public aBaseClass
{
public:
    void testFunction(){cout<<"hello derived one";}
};

class anotherDerivedClass:public aDerivedClass
{
public:
    void testFunction(){cout<<"hello derived two";}

};
int main()
{
    aBaseClass *aBaseClassPointer;
    aBaseClassPointer=new aDerivedClass;
    aBaseClassPointer->testFunction();
}

The above code does not support run time polymorphism. Lets run and analyze it. The output is

hello base

Just change the line void testFunction(){cout<<"hello base";} to virtual void testFunction(){cout<<"hello base";} in aBaseClass. Run and analyze it. We see that runtime polymorphism is achieved. The calling of appropriate function is determined at run time.

Again change the line aBaseClassPointer=new aDerivedClass to aBaseClassPointer=new anotherDerivedClass in main function and see the output. Thus the appropriate function calling is determined at run time (when the program is running).

score 0 · Answer 6 · answered Aug 11 '10 at 09:52

You need to have some useful business method declared in the base and in each derived class. Then you have code such as

 a->someMethod();

Now the a pointer might point to an instance of any of the derived classes, and so the type of what a is pointing to must determine which someMethod() is called.

What exactly does runtime polymorphism mean?

6 Answers6

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