C++ 在 std::move() 之后 unique_ptr 会发生什么？
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What happens to unique_ptr after std::move()?
提问by user3496846
This code is what I want to do:
这段代码是我想要做的：
Tony& Movie::addTony()
{
Tony *newTony = new Tony;
std::unique_ptr<Tony> tony(newTony);
attachActor(std::move(tony));
return *newTony;
}
I am wondering if I could do this instead:
我想知道我是否可以这样做：
Tony& Movie::addTony()
{
std::unique_ptr<Tony> tony(new Tony);
attachActor(std::move(tony));
return *tony.get();
}
But will *tony.get()be the same pointer or null? I know I could verify, but what is the standard thing for it to do?
但是会*tony.get()是同一个指针还是null？我知道我可以验证，但它的标准做法是什么？
回答by Benjamin Lindley
No, you cannot do that instead. Moving the unique_ptrnulls it. If it didn't, then it would not be unique. I am of course assuming that attachActordoesn't do something silly like this:
不，你不能这样做。移动unique_ptr空值它。如果没有，那么它就不是唯一的。我当然假设这attachActor不会做这样愚蠢的事情：
attachActor(std::unique_ptr<Tony>&&) {
// take the unique_ptr by r-value reference,
// and then don't move from it, leaving the
// original intact
}
Section 20.8.1 paragraph 4.
第 20.8.1 节第 4 段。
Additionally, u (the unique_ptr object) can, upon request, transfer ownership to another unique pointer u2. Upon completion of such a transfer, the following postconditions hold:
-- u2.p is equal to the pre-transfer u.p,
-- u.p is equal to nullptr, and
-- if the pre-transfer u.d maintained state, such state has been transferred to u2.d.
此外，u（unique_ptr 对象）可以根据请求将所有权转移到另一个唯一指针 u2。完成这样的传输后，以下后置条件成立：
-- u2.p 等于传输前 up，
-- up 等于 nullptr，并且
-- 如果传输前 ud 保持状态，则该状态已转移到u2.d。
回答by 5gon12eder
The standard says (§ 20.8.1.2.1 ? 16, emphasis added) that the move constructor of std::unique_ptr
标准说（§ 20.8.1.2.1 ? 16，强调添加）移动构造函数 std::unique_ptr
unique_ptr(unique_ptr&& u) noexcept;Constructs a
unique_ptrby transferring ownershipfromuto*this.
unique_ptr(unique_ptr&& u) noexcept;
unique_ptr通过将所有权从转移到u来构造*this。
Therefore, after you move-construct the temporary object that gets passed as argument to attachActorform your tony, tonyno longer owns the object and hence tony.get() == nullptr. (This is one of the few cases where the standard library actually makes assertions about the state of a moved-away-from object.)
因此，后移动，构建被作为参数传递给临时对象attachActor形成的tony，tony不再拥有的对象，因此tony.get() == nullptr。（这是标准库实际上对远离对象的状态做出断言的少数情况之一。）
However, the desire to return the reference can be fulfilled without resorting to naked newand raw pointers.
但是，无需诉诸裸new指针和原始指针即可满足返回引用的愿望。
Tony&
Movie::addTony()
{
auto tony = std::make_unique<Tony>();
auto p = tony.get();
attachActor(std::move(tony));
return *p;
}
This code assumes that attachActorwill not drop its argument on the floor. Otherwise, the pointer pwould dangle after attachActorhas returned. If this cannot be relied upon, you'll have to re-design your interface and use shared pointers instead.
此代码假定attachActor不会将其参数放在地板上。否则，指针p会在attachActorhas returned之后悬空。如果不能依赖，则必须重新设计界面并改用共享指针。
std::shared_ptr<Tony>
Movie::addTony()
{
auto tony = std::make_shared<Tony>();
attachActor(tony);
return tony;
}
