C ++将指针的指针设置为null吗？

Question

I'm working on some kind of smart pointer technique but there is one piece I'm missing. I tried several combinations but the logic is as follow:

UInt *obj = new UInt;
UInt *ref;
ref = obj;

delete obj;
obj = NULL;

if (ref == NULL)
{
    // It works
}
else
{
    // It failed
}

Is there any way to hit "It Works" without setting ref to NULL explicitly?

EDIT:

A more appropriate scenario would be something like this:

class A
{
public:

    A(): ref(NULL) {}
    ~A()
    {
        if (ref != NULL)
            delete ref;
    }
    int *ref;
};

    int *obj = new int;
    A *host = new A();

    host->ref = obj; ???

    delete obj;
      obj = NULL;

    if (host->ref == NULL)
    {
        // It works.
    }
    else
    {
        // It failed.
    }

...

Can't use int*& ref as a class member though.... must be close.

Answer 1

As you say, you should be using a smart pointer:

#include <memory>

std::shared_ptr<UInt> obj = std::make_shared<UInt>();
std::weak_ptr<UInt> ref = obj;

obj.reset();

if (ref.expired())
{
    // It works
}
else
{
    // It failed
}

Don't try managing your own memory when the standard library has facilities to do it for you.

Answer 2

Declare ref as a reference to pointer

Uint*& ref = obj;

ref will now refer to the obj pointer.

Answer 3

Intially both obj and ref point to the (same) UInt instance. Then you delete the instance and set obj to NULL . But ref is just like any other regular pointer, so it still points to the deleted instance.

Instead you can create a reference variable, or in this case a 'reference pointer' by adding & to the declaration:

Uint*& ref = obj;

Then ref really refers to obj and keeps the same value (pointer) as obj .