Question

Consider the below code snippet:

int main()
{
    const int i=3;
    int *ptr;

    ptr=const_cast<int*>(&i);
    *ptr=5;

    cout<<"i= "<<i<<endl;  <------------------- statement 1
    cout<<"*ptr= "<<*ptr<<endl;  <------------- statement 2

    return 0;
}

I am getting the output as:

i= 3
*ptr= 5

http://ideone.com/Bvme6

Why is the value of i is not changed through pointer?

I know casting away the const-ness of a variable which is explicitly declared as const and modifying its value is ‘Undefined Behavior’. I am curious to know: Is it any compiler optimization mechanism that ‘compiler replaces the variable in the program with the value’?.
It means the statement 1 is interpreted by the compiler as:

cout<<"i= "<<3<<endl;

Even if the statement

ptr=const_cast<int*>(&i);    

is replaced by

 ptr=(int*)(&i);

I am getting the same output: http://ideone.com/5lzJA

Answer 1

Is it any compiler optimization mechanism that compiler replaces the variable in the program with the value?

Yes; that will be why you don’t see the value changing. The behaviour of trying to modify a const object is left undefined in order to allow optimisations like that (as well as allowing the objects to be placed in unwritable memory).