C++ smart pointers address

Question

I am a little confused with smart pointers. In the following code, should the & operator return the adress of the smart pointer allocation or the address of the pointer it's controlling?

main() {
    std::shared_ptr<int> i = std::shared_ptr<int>(new int(1));
    std::shared_ptr<int> j = i;
    printf("(%p, %p)\n", &i, &j);
}

Running the code, I got different address. If I run an equivalent code with raw pointers, I get the same adress:

main() {
    int e = 1;
    int *k = &e;
    int *l = k;

    printf("(%p, %p)\n",k,l);
}

use i.get() and j.get() rather than &i and &j to get the pointer held by the smart pointer. — Laserallan, May 24 '13 at 04:38
@Laserallan: That is the answer.. can you post that as answer? — Asha, May 24 '13 at 04:40

score 14 · Accepted Answer · answered May 24 '13 at 04:50

In the first example, you're getting the address of the smart pointer object. The raw pointer contained within a smart pointer is provided via the get() function.

The address-taking of smart pointers works almost exactly the same as regular pointers, actually. The raw pointer equivalent of your first example would be this:

main() {
    int e = 1;
    int *k = &e;
    int *l = k;

    printf("(%p, %p)\n",&k,&l); // now they're different!
}

And the smart pointer equivalent of your second example would be this:

main() {
    std::shared_ptr<int> i = std::shared_ptr<int>(new int(1));
    std::shared_ptr<int> j = i;
    printf("(%p, %p)\n", i.get(), j.get()); // now they're the same!
}

Thanks for the answer. But, is the -> overloaded in shared pointer? Cause I can do, for example i->someMethod() and no (i.get())->someMethod()... — Lucas Kreutz, May 24 '13 at 04:55

score 0 · Answer 2 · answered May 24 '13 at 04:40

0

Please call the .get() member function of std::shared_ptr to get the the address you want.

answered May 24 '13 at 04:40

mumu

197
7

fatihk · Answer 3 · 2013-05-24T05:52:55.680

0

Here, the main trick is that equality operator (=) for shared pointers are defined in such a way that when you do:

std::shared_ptr<int> j = i;

j will not be a complete copy of i. but it will just keep the same raw pointer the shared pointer i holds and therefore, their addresses will be different.

edited May 24 '13 at 05:52

answered May 24 '13 at 05:47

fatihk

7,789
1
26
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C++ smart pointers address

3 Answers3