vector.push_back右值和复制删除 [英] vector.push_back rvalue and copy-elision

问题描述

我将这样的临时对象 push_back 转换成 vector ，

I push_back a temporary object into a vector like this,

vector<A> vec;
vec.push_back(A("abc"));

编译器将使用复制删除来构造临时 A（ abc ）直接放入 vector 中，这样 A 的副本ctor不会将临时对象推入 vec 时被触发。

will the compiler apply copy-elision to construct the temporary A("abc") directly into the vector, so that A's copy ctor won't be triggered when pushing the temporary object into vec.

推荐答案

如果有一个支持rvalue引用的编译器，它将被移到向量中，这有时很便宜。

If you have a compiler that supports rvalue references, it will be moved into the vector, which is sometimes quite cheap.

一种替代方法是直接在向量中构造对象，可以使用 vec.emplace_back（ abc）; 完成。

An alternative to that is to directly construct the object in the vector, which can be done with vec.emplace_back("abc");. This only invokes one constructor.

这两个都是C ++ 11特性。此处不允许进行复制省略，因此如果没有这些功能，仍将复制。

Both of these are C++11 features. Copy elision is not allowed here, so without those features the copy will still be made.

但是，如果复制构造函数没有明显的副作用（不应该这样做）。无论如何，智能编译器仍然可以执行这种优化，因为按原样规则允许进行任何优化，从而产生相同的可观察行为。不过，我不知道当前是否有编译器会这样做。如果不是这样，我怀疑有人会花精力添加这种优化，因为右值引用消除了这种需要。

However, if the copy constructor has no observable side-effects (which it shouldn't have anyway), a smart compiler may still perform that optimisation, because the "as-if" rule allows any optimisation that results in the same observable behaviour. I don't know if any current compiler does that, though. If not, I doubt anyone will spend the effort to add such an optimisation because rvalue references put an end to that need.

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