Question

I understood why a base class pointer is made to point to a derived class object. But, I fail to understand why we need to assign to it, a base class object, when it is a base class object by itself.

Can anyone please explain that?

#include <iostream>
using namespace std;
class base {
     public:
        virtual void vfunc() {
            cout << "This is base's vfunc().\n";
         }
};
 class derived1 : public base {
      public:
            void vfunc() {
                 cout << "This is derived1's vfunc().\n";
    }
};
int main()
{
     base *p, b;
     derived1 d1;
     // point to base
     p = &b;
     p->vfunc(); // access base's vfunc()
     // point to derived1
     p = &d1;
     p->vfunc(); // access derived1's vfunc()
     return 0;
}

Answer 1

Because pointers by themselves cannot do anything.
A pointer has to point to a valid object so that you can make any use of it.

Why the above statement?

A step by step explanation will perhaps clear your doubt.

Step 1:

base *p;

Creates a pointer p which can store the address of an object of class base. But it is not initialized it points to any random address in memory.

Step 2:

p = &b;

Assigns the address of an valid base object to the pointer p. p now contains the address of this object.

Step 3:

p->vfunc(); // access base's vfunc() 

Dererences the pointer p and Calls the method vfunc() on the object pointed by it. i.e: b.

If you remove the Step 2:, Your code just tries to Dereference a Uninitialized pointer and would cause a Undefined Behavior & most likely a crash.