const_cast usage.

Definition :It allows the programmer to strip the const qualifier, making any object modifiable.


Lets start with an example :

const int a = 10;
// ++a; Compiler error as const value can't be changes.
int *ptr = const_cast&a;


*ptr = 80;


cout<<a<<" == "<<*ptr;

OutPut : 10 == 80.

Surprised ?
Value of a is not changed?
Then what is the use of const_cast?

Lets do some analysis : Start with the address of both variables. May be they point to different locations in memory. Thats why the values they contains are different.

if (&a == ptr) {
cout<<"Addresses are same";
}
else {
cout<<"Different addresses";
}

Now the OutPut is :
Addresses are same.

Big surprise : If the addresses are same then how can be value at them are different. How two different values can be reside on the same memory locations.

Here is the reason :
const_cast is remove the constness of the object. This statement is true here too but not visible due to compiler optimization. Actually when we declare any variable as a const then compiler take it as literal. So when ever it looks for "a" in given example it is replaced by 10. That is : After compilation statement look like :

cout<< 10<<" == "<<*ptr;

Even if we use the address operator, the compiler is being smart enough to realize that (*&x) is the same expression as (x), so it's optimizing away the indirection. For Example :


cout<<*(&a)<<" =="<<*ptr;

But the result is same in this case too : 10 == 80.

How can be know that the value at this address is changed or not?
Answer of this question is simple program to try yourself :

#include

#include

#include "Main.h"

#include "C.h"

using namespace std;

voidPrintValue(const int*ptr)

{

      cout<<*ptr;

}

void main()

{

      const inti = 10;

      int *k = const_cast<int*>(&i);

      *k = 80;

      cout<<&i<<" == "<<k<<endl;

      cout<<*&i<<" != "<<*k<<endl;

      PrintValue(&i);

      cout<<" == "<<*k;

      _getch();

}

Check the output now.

Thanx