C++“移出"容器

C++ quot;move fromquot; container(C++“移出容器)

问题描述

在 C++11 中，当我们想要将值移动(破坏性复制)到容器中时，我们可以使用 std::move 来提高效率:

In C++11, we can get an efficiency boost by using std::move when we want to move (destructively copy) values into a container:

SomeExpensiveType x = /* ... */;
vec.push_back(std::move(x));

但我找不到任何相反的方向.我的意思是这样的:

But I can't find anything going the other way. What I mean is something like this:

SomeExpensiveType x = vec.back(); // copy!
vec.pop_back(); // argh

这在适配器的 stack 上更为频繁(复制弹出).这样的事情是否存在:

This is more frequent (the copy-pop) on adapter's like stack. Could something like this exist:

SomeExpensiveType x = vec.move_back(); // move and pop

为了避免复制?这已经存在了吗?我在n3000中找不到类似的东西.

To avoid a copy? And does this already exist? I couldn't find anything like that in n3000.

我有一种感觉我错过了一些非常明显的东西(比如它的不必要性)，所以我为ru dum"做好了准备.:3

I have a feeling I'm missing something painfully obvious (like the needlessness of it), so I am prepared for "ru dum". :3

推荐答案

我可能完全错了，但这不是你想要的

I might be total wrong here, but isn't what you want just

SomeExpensiveType x = std::move( vec.back() ); vec.pop_back();

假设 SomeExpensiveType 有一个移动构造函数.(对于您的情况显然是正确的)

Assuming SomeExpensiveType has a move constructor. (and obviously true for your case)

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