C++ set a pointer of pointer to null?

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.

Is `UInt` your smart pointer class? Smart pointers should not be allocated on the heap. — Seth Carnegie, Feb 01 '13 at 16:37
What does "pointer of pointer" stand for in the title of the question? — AnT stands with Russia, Feb 01 '13 at 16:38
I'm sorry, forget about the smart pointer scenario. UInt could be int instead. — neosettler, Feb 01 '13 at 17:06
yeah i am going to have to concur with the other comments. those pointers seem pretty dumb to me. — thang, Feb 01 '13 at 17:07
You are right, I did not try to illustrate a smart pointer technique but the missing piece I need to make it work. My bad. — neosettler, Feb 01 '13 at 17:20

score 4 · Answer 1 · answered Feb 01 '13 at 16:47

4

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.

answered Feb 01 '13 at 16:47

ecatmur

152,476
27
293
366

Exactly, so that people who read your code will not need to read and understand your own implementation, when there is a standard that everybody knows. Also, most probably, standard will have it better than yours. – Etherealone Feb 01 '13 at 16:50
True, I never use the std lib, I guess I could give it a try. I'm still interested how I could accomplish the example. – neosettler Feb 01 '13 at 17:10

score 3 · Answer 2 · answered Feb 01 '13 at 16:39

3

Declare ref as a reference to pointer

Uint*& ref = obj;

ref will now refer to the obj pointer.

answered Feb 01 '13 at 16:39

Andreas Brinck

51,293
14
84
114

score 1 · Answer 3 · answered Feb 01 '13 at 16:41

1

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.

answered Feb 01 '13 at 16:41

Veger

37,240
11
105
116

Oh excellent, I used this example for simplicity sake but what if ref is a class member? – neosettler Feb 01 '13 at 17:13
http://stackoverflow.com/questions/892133/should-i-prefer-pointers-or-references-in-member-data provides some details, cons and pros of references as class members vs pointers – Veger Feb 01 '13 at 20:18

C++ set a pointer of pointer to null?

3 Answers3