If we assign a dynamically allocated pointer to another pointer, what happens?-CodePudding

#include <iostream>
using namespace std;

int main()
{
    int* A = new int[2];
    A[0] = 10000;
    A[1] = 2;
    int*B = A;
    delete B;
    cout << B << A;
}

I found the code above, delete B; removes the data of array A. Can we remove allocated memory by free some another assigned pointer?

CodePudding user response：

You don't delete pointers. You delete the thing pointed to.

A, in your example isn't an array. It is a pointer that points to the first element of an array. When you write int* B = A;, you make B a pointer to the same memory. At that point A and B are identical, and the array they point to can be freed via either of them.

It doesn't matter how many times you copy a pointer. All that matters is that the thing pointed to by the pointer you pass wo your deletion expression was allocated with the corresponding allocation expression (i.e. new/delete, new[]/delete[], malloc/free, etc).

CodePudding user response：

Can we remove allocated memory by free some another assigned pointer?

Yes, as long as you do it correctly i.e., using the appropriate delete-expression.

From new.delete.single-12:

Requires: ptr shall be a null pointer or its value shall represent the address of a block of memory allocated by an earlier call to a (possibly replaced) operator new(std::size_t) or operator new(std::size_t, std::align_val_t) which has not been invalidated by an intervening call to operator delete.

This means that in your program, we are allowed to write:

delete [] B;

On the other hand delete B; is incorrect because we must use array-form of delete-expression here.