为什么`std :: initializer_list`经常通过值传递？ [英] why is `std::initializer_list` often passed by value?
问题描述
在我在SO上看到的几乎每个帖子中，涉及到一个 std :: initializer_list ，人们倾向于传递 std :: initializer_list 按值。根据本文：
In almost every post I see on SO, involving a std::initializer_list, people tend to pass a std::initializer_list by value. According to this article:
http://cpp-next.com/archive/2009/08/want-speed-pass-by-value/
pass by value，如果想要创建传递对象的副本。但是复制 std :: initializer_list 不是一个好主意，因为
one should pass by value, if one wants to make a copy of the passed object. But copying a std::initializer_list is not a good idea, as
a
std :: initializer_list不复制基础对象。
在
的生命周期结束后，底层数组不能保证存在。
Copying a
std::initializer_listdoes not copy the underlying objects. The underlying array is not guaranteed to exist after the lifetime of the original initializer list object has ended.
所以为什么它的一个实例经常通过值而不是通过 const& ，这保证不会造成不必要的副本？
So why is an instance of it often passed by value and not by, say const&, which guaranteed does not make a needless copy?
推荐答案
它是通过价值，因为它很便宜。 std :: initializer_list 作为一个thin wrapper，最有可能被实现为一对指针，因此复制与通过引用传递一样便宜。此外，我们实际上并不执行副本，我们（通常）执行移动，因为在大多数情况下，参数是从临时建立的。但是，这不会对性能产生影响 - 移动两个指针和复制它们一样昂贵。
It’s passed by value because it’s cheap. std::initializer_list, being a thin wrapper, is most likely implemented as a pair of pointers, so copying is (almost) as cheap as passing by reference. In addition, we’re not actually performing a copy, we’re (usually) performing a move since in most cases the argument is constructed from a temporary anyway. However, this won’t make a difference for performance – moving two pointers is as expensive as copying them.
另一方面，复制的元素可能会更快 ，因为我们避免一个额外的解引用（引用的）。
On the other hand, accessing the elements of a copy may be faster since we avoid one additional dereferencing (that of the reference).
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