C++ Pointer Confusion

Question

Please Explain the following code

#include <iostream>

using namespace std;

int main()
{
    const int x = 10;
    int * ptr;
    ptr = (int *)( &x );    //make the pointer to constant int*
    *ptr = 8;               //change the value of the constant using the pointer.
    //here is the real surprising part
    cout<<"x: "<<x<<endl;          //prints 10, means value is not changed
    cout<<"*ptr: "<<*ptr<<endl;    //prints 8, means value is changed
    cout<<"ptr: "<<(int)ptr<<endl; //prints some address lets say 0xfadc02
    cout<<"&x: "<<(int)&x<<endl;   //prints the same address, i.e. 0xfadc02
    //This means that x resides at the same location ptr points to yet 
    //two different values are printed, I cant understand this.

    return 0;
}

Answer 1

*ptr = 8;

This line causes undefined behavior because you are modifying the value of a const qualified object. Once you have undefined behavior anything can happen and it is not possible to reason about the behaviour of the program.

Answer 2

Since x is a const int , the compiler will most likely, in the places where you use x , directly substitute the value that you've initialized with (where possible). So the line in your source code:

cout<<"x: "<<x<<endl;

will be replaced by this at compile time:

cout<<"x: "<<10<<endl;

That's why you still see 10 printed.

But as Charles explained, the behaviour is undefined, so anything could happen with code like this.