Pure virtual functions and virtual table

A pure virtual function is a virtual function whose declaration ends in =0:

class Base {
  // ...
  virtual void f() = 0;
  // ...

Quote:

what values will be stored in virtual table if the class has pure virtual function

There is a subtle difference between pure- and non-pure-functions, take a look:

Quote:

In the case of non-pure virtual functions, each entry in the vtable will refer to the final-overrider or a thunk that adapts the this pointer if needed.
In the case of a pure-virtual function, the entry in the vtable usually contains a pointer to a generic function that complains and aborts the program with some sensible message (pure virtual function called within this context or similar error message).

Quote:

what does 0 indicates in pure virtual functions

A 0 indicates that this is a pure virtual function! A pure virtual function implicitly makes the class it is defined for abstract (unlike in Java where you have a keyword to explicitly declare the class abstract). Abstract classes cannot be instantiated. Derived classes need to override/implement all inherited pure virtual functions. If they do not, they too will become abstract.

[EDITED]

Quote:

1) How does the compiler knows whether vtable has the pure virtual function entry or virtual function .

2) How we can make the call to base class pure virtual function. I tried to make a static call from derived class i got a linker error? .If possible can you please provide sample code

To 1): The compiler don't need to know! In case of pure-virtual-function is a NULL instead of address, it raises the exception. In case of a regular virtual function there is an address of the implemented function, it will be called.

To 2):

class A
{
public: 
    virtual void foo()=0;
};

class B : public virtual A
{
public:
    void foo() { cout<<"foo()"; }
};

int main()
{
    A* obj = new B();
    obj->foo();
    return 0;
}