移动会使对象处于可用状态吗？
假设我有两个向量，我将一个向量移动到另一个向量，v1 = std::move(v2);此后v2仍处于可用状态吗？
Say I have two vectors and I move one unto the other, v1 = std::move(v2); will v2 still be in a usable state after this?
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从 n3290，17.6.5.15 库类型的移出状态 [lib.types.movedfrom]
由于状态有效，这意味着您可以安全地对 v2 进行操作（例如，通过分配给它，这会将其恢复到已知状态）。然而，由于它是未指定的，这意味着只要它处于这种状态，您就不能依赖 v2.empty() 的任何特定值（但调用它不会使程序崩溃）。
请注意，移动语义的这条公理（“从对象中移动的对象保持有效但未指定的状态”）是所有代码（大多数时候）都应该努力实现的目标，而不仅仅是标准库组件。与复制构造函数的语义非常相似，应该进行复制，但不强制这样做。
From n3290, 17.6.5.15 Moved-from state of library types [lib.types.movedfrom]
Since the state is valid, this means you can safely operate on
v2(e.g. by assigning to it, which would put it back to a known state). Since it is unspecified however, it means you cannot for instance rely on any particular value forv2.empty()as long as it is in this state (but calling it won't crash the program).Note that this axiom of move semantics ("Moved from objects are left in a valid but unspecified state") is something that all code should strive towards (most of the time), not just the Standard Library components. Much like the semantics of copy constructors should be making a copy, but are not enforced to.
不，它处于未指定的状态。
摘自open-std-org文章-
No, it is left in an unspecified state.
Excerpt from open-std-org article -
如果您想要在移动后使用v2，您将需要执行以下操作：
此时，v1 将具有 v2 的原始内容，并且 v2 将处于明确定义的空状态。这适用于所有 STL 容器（以及字符串），如果您正在实现自己的支持移动语义的类，您可能会想要做类似的事情。
如果您的 STL 的特定实现实际上确实使对象处于空状态，那么第二个 clear() 本质上将是无操作。事实上，如果是这种情况，编译器在移动后消除clear()将是合法的优化。
If you want to use v2 after the move, you will want to do something like:
At this point, v1 will have the original contents of v2 and v2 will be in a well-defined empty state. This works on all of the STL containers (and strings, for that matter), and if you are implementing your own classes that support move-semantics, you'll probably want to do something similar.
If your particular implementation of the STL actually DOES leave the object in an empty state, then the second clear() will be essentially a no-op. In fact, if this is the case, it would be a legal optimization for a compiler to eliminate the clear() after the move.