Reinitialize object without invalidating pointers

[Xriuk]

I have:

MyClass* A = new MyClass(1);
MyClass* B = A;
/* doing things with A */
// Here I need to reinitialize A (eg. new MyClass(2)) and update B automatically 

Is this possible? Or how can this problem be solved?

[Stalker]

Yes it is possible. You can pass another new address in 'A' without losing the data in 'B'. That is one of the advantages of pointers.

[LB]

Why are you using pointers in the first place? Why do you need two pointers pointing to the same object in the same scope?

[Moschops]

MyClass* A = new MyClass(1);
MyClass* B = A;
A = new MyClass(2);

You still have the original object (because B is pointing to it) and you now also have this new object (because A is pointing to it).

[Xriuk]

I was saying if it was possible by changing A to update B too. Sorry for the misleading.

[codekiddy]

if value of A is modified then value of B is updated since they point to same location.

if address of A is changed then address of B needs to be deleted and reassigned to new address of A.
addresses can't be automatically updated.

however you can use std::shared_ptr and invoke reset() on A which will reduce the work needed.

for example:

#include <memory>
#include <iostream>
using namespace std;

class MyClass
{
public:
	MyClass(int x) : x(x) {}
	int x;
};

int main()
{
	auto A = make_shared<MyClass>(1);
	auto B = A;
	cout << B->x << endl; // prints 1

	A.reset(new MyClass(2)); // assign new address to A
	B = A; // delete old address and re-assign B to A

	cout << B->x << endl; // prints 2

	return 0;
}

[Moschops]

If you want B to basically be just another name for A, which is what it sounds like, you're asking about references.

MyClass* A = new MyClass(1);
MyClass*& B = A;

Now, B is just another name for A. Anything that happens to A happens to B as well, because B is just another name for A.