Is there a recommended way to test if a smart pointer is null?

Question

I'm trying to check if a std::shared_ptr is null. Is there a difference between doing

std::shared_ptr<int> p;
if (!p) { // method 1 }
if (p == nullptr) { // method 2 }

Solution

Is there a difference between doing

 std::shared_ptr<int> p;
 if (!p) { // method 1 }
 if (p == nullptr) { // method 2 }

No, there's no difference. Either of that operations has a properly defined overload.

Another equivalent would be

 if(p.get() == nullptr)

Comments

[Richard Hodges]

No difference at all. You could also write if(not p.get())

[πάντα ῥεῖ]

What about if (p.get() == nullptr)?

[Galik]

No difference. One is a lot less typing.

[Yakk - Adam Nevraumont]

In theory the 3rd one could be more complex in relatively insane situations, as it must get the T* prior to comparing to nullptr. Imagine a type erased smart pointer that must call a vtable method prior to getting the T*, but if the value is null guarantees a nullptr vtable. Method 1 and 2 can short circuit the possible extract-T*, while 3 could not, because .get() does not know it will be just checked for null.

[Sergey Kolesnik]

I do not agree with your intent being explicit in your example. The intent is to check for nullptr, so p != nullptr is more explicit for the question. In your example you are only explicit about your intent to invoke operator bool.

[Antonio]

@SergeyKolesnik The intent is about knowing if the shared_ptr has been set/reset