CODEWITHSUNDEEP
c++pointersintconstants
AkiyamaKunka
AkiyamaKunka

Reputation: 21

Why the two const int objects have same address but not same value after "const" cast?

Please see, I defined const int & a = 1 by reference, and forced the pointer to convert pa to a, and changed the value of a by changing pa. This was successful.

But when const int k = 1 is defined and the above operation is repeated, although pk and k are the same address, * pk and k are not the same value.

What is the principle behind it?

Could you please explain how the IDE work with the memory allocation when I do so?

const int &a = 1;
    int *pa = (int*)&a;
    cout << &a << endl;
    cout << pa << endl;
    *pa = 2;
    cout << a << endl;
//And here is the outcome.


//0x7ffeeb5d8a24
//0x7ffeeb5d8a24
//2

So here we changed a successfully.


    const int k = 1;
    cout << &k << endl;
    int *pk = (int*)&k;
    cout << &k << endl;
    cout << pk << endl;
    *pk = 2;
    cout << *pk << ' ' << k;

//0x7ffeeb5d8a14
//0x7ffeeb5d8a14
//0x7ffeeb5d8a14
//2 1
//Process finished with exit code 0

*pkandkhere have same address, but not the same value! How could this happen?

Upvotes: 0

Views: 92

Answers (1)

Remy Lebeau
Remy Lebeau

Reputation: 596582

This is technically undefined behavior. It is not safe to modify const data at runtime, even if you cast away its const-ness. Only data that is initially non-const and has const added to it later can safely have that const casted away.

But, to explain why you are seeing the output that is confusing you - the answer is compiler optimization!

The compiler sees const int k = 1; as a compile-time constant and assumes k will never change value (and rightly so!). When the compiler sees cout << *pk << ' ' << k;, it is allowed to replace k with the value 1 right there at the call site at compile-time, thus you are really executing cout << *pk << ' ' << 1; regardless of what you do to k at runtime. But, being that k is a compile-time constant that you take the address of, it has to be stored in memory, and the compiler is free to (and likely will) store that value in read-only memory. Altering such memory at runtime is likely to crash your app, but that is not a guarantee.

Upvotes: 5

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