CODEWITHSUNDEEP
c++smart-pointersc++17
davidhigh
davidhigh

Reputation: 15498

C++ why does `observer_ptr<W>` have a constructor taking `W*`?

As far I understood, the proposed std::observer_ptr is related to std::unique_ptr in the same way as std::weak_ptr is related to std::shared_ptr.

So why does the std::observer_ptr<W> interface, according to the proposal N4282, allow the construction from a W* pointer?

This implies an implementation which contains a W* as member, maybe similar to the pseudo-implementations given in this answer, which most simply proposes

template<typename T>
using observer_ptr = T*;

As a consequence, this seems to outrule validity checks as in the following:

std::unique_ptr<W> u = std::make_unique<W>();
std::observer_ptr<W> o(uptr.get());
uptr.reset();

if(o)
{
     //u is already nullptr, but o doesn't know
     o->foo();  //invalid dereferentation
}

Instead I would expect to only be allowed to do the following:

std::unique_ptr<W> u = std::make_unique<W>();
std::observer_ptr<W> o(uptr);
uptr.reset();

if(o)
{
     //execution doesn't get here, o is nullptr
}

This is equivalent to what one can do with std::weak_ptr by locking it, and is imo the central advantage observer_ptr could offer over a non-owning raw pointer.

So, again, why isn't it enforced?

Upvotes: 0

Views: 351

Answers (1)

juanchopanza
juanchopanza

Reputation: 227468

As far I understood the proposed std::observer_ptr is related to std::unique_ptr in the same way as std::weak_ptr is related to std::shared_ptr.

That is a misconception. It has no relation to unique_ptr. It expresses that there is no connection to the ownership of the pointee.

Upvotes: 9

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