为什么/何时应该在std :: vector <>上使用std :: unique / shared_ptr（std :: vector <>）？

Question

I'm a little bit confused about the main use of std::unique/shared_ptr(std::vector<>) when I can simply use a std::vector<> , which, as I know, is itself inherently a dynamic array. As I have also seen around, people say that there is no any performance difference between these two. So, based on all this, what is the point of using a smart pointer pointing to a container (in this case, a vector) instead of a vector alone?

Answer 1

First of all, you shouldn't be using std::shared_ptr unless you need the specific "shared ownership" semantics associated with std::shared_ptr . If you need a smart pointer, you should default to std::unique_ptr by default, and only switch away from it in the scenario where you expressly find that you need to.

Secondly: ostensibly , the reason to prefer std::unique_ptr<TYPE> over TYPE is if you plan to move the object around a lot. This is the common design paradigm for large objects that are either unmovable, or otherwise expensive to move—ie they implemented a Copy Constructor and didn't implement a Move Constructor, so moves are forced to behave like a Copy.

std::vector , however, does have relatively efficient move semantics: if you move a std::vector around, regardless of how complex its contained types are, the move only constitutes a couple of pointer swaps. There's no real risk that moving a std::vector will incur a large amount of computational complexity. Even in the scenario where you're overriding a previously allocated array (invoking the Destructors of all objects in the vector), you'd still have that complexity if you were using std::unique_ptr<std::vector<TYPE>> instead, saving you nothing.

There are two advantages to std::unique_ptr<std::vector<TYPE>> . The first of which is that it gets rid of the implicit copy constructor; maybe you want to enforce to maintaining programmers that the object shouldn't be copied. But that's a pretty niche use. The other advantage is that it allows you to stipulate the scenario where there's no vector, ie vec.size() == 0 is a different condition than doesNotExist(vec) . But even in that scenario, you should be preferring std::optional<std::vector> instead, which better conveys through the code the intent of the object. Granted, std::optional is only available in C++17→ Code, so maybe you're in an environment that hasn't implemented it yet. But otherwise, there's little reason to use std::unique_ptr<std::vector> .

So in general, I don't believe there are practical uses for std::unique_ptr<std::vector> . There's no practical performance difference between it and std::vector , and using it will just make your code needlessly complex.