Why Does Modifying a `const` Variable's Memory Address Show Different Values?

Understanding Memory Address Behavior with Variables and Constants

In a code snippet involving a const variable and a pointer to it, the observation of different values at the same memory address raises questions.

Code Analysis

#include <iostream>
using namespace std;

int main() {
    const int N = 22;
    int * pN = const_cast<int*>(&N);
    *pN = 33;
    cout << N << '\t' << &N << endl;
    cout << *pN << '\t' << pN << endl;
}

Output

22      0x22ff74

33      0x22ff74

Question

Why do we observe two different values (22 and 33) at the same memory address (0x22ff74)?

Answer

Contrary to the observation, there are not two different values stored at the same memory address. This is a result of compiler optimizations.

The compiler has the authority to treat any mention of a const variable as if the variable's compile-time value is used directly. In this code, the compiler views the code as:

int * pN = const_cast<int*>(&22);
*pN = 33;

The compiler is within its rights to apply this optimization. However, it's important to note that the compiler is not limited to this action. It may employ other optimizations, including ones that could potentially erase data on your hard drive if you engage in risky practices like modifying memory associated with const variables.

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