取消引用分配给双倍增量的OutputIterator [英] Dereference-assignment to a doubly incremented OutputIterator

问题描述

根据（优秀）问题 C ++ OutputIterator增量后要求，我们观察对于可解除引用且可递增的值 r of OutputIterator type X 和值 o 适当的类型，表达式

Per the (excellent) question C++ OutputIterator post-increment requirements, we observe that for a dereferenceable and incrementable value r of OutputIterator type X, and value o of appropriate type, the expression

*r++ = o;

有效且具有与

X a(r);
++r;
*a = o;

但是， a 如果 r 在中间期间增加了多次，则取消引用 - 可分配;也就是说，这段代码有效吗？

However, is it still the case the a is dereference-assignable if r has been incremented more than once in the intervening period; that is, is this code valid?

X a(r);
++r;
++r;
*a = o;

很难看出某个值的操作如何影响另一个值的操作有效性但是，例如 InputIterator （24.2.3）在 ++ r的后置条件下：

It's difficult to see how operations on a value can have an effect on the validity of operations on another value, but e.g. InputIterator (24.2.3) has, under the postconditions of ++r:

先前值 r 的任何副本都不需要

dereferenceable或
域 == 。

Any copies of the previous value of r are no longer required either to be dereferenceable or to be in the domain of ==.

相关章节： 24.2.2迭代器， 24.2.4输出迭代器， 17.6.3.1模板参数要求。

此外，如果不要求有效，是否有任何情况下利用其无效性将有助于实施（wrt效率，简单性） OutputIterator 类型，同时仍然遵守现有要求？

Also, if this is not required to be valid, are there any situations where exploiting its non-validity would aid in the implementation (w.r.t. efficiency, simplicity) of an OutputIterator type while still observing the existing requirements?

推荐答案

该问题于2004年提出，因为缺陷485 ，以及 n3066 澄清了这个问题，要求输出迭代器只需要支持交替增量和解除引用/分配的顺序。因此，在您的示例中， r 在第一个 ++ r 之后无需递增，除非存在干预取消引用/分配。 SGI的 STL 也需要此行为（请参见脚注3）。如上所述，n3225没有修复n3066，所以缺陷2035 被提出;但唉，修复版没有进入C ++ 11的发布版本（ISO / IEC 14882：2011）。

This issue was raised in 2004 as defect 485, and the wording in n3066 clarifies the issue, requiring that an output iterator need only support a sequence of alternating increments and dereference/assignments. So in your example, r need not be incrementable after the first ++r, unless there is an intervening dereference/assignment. This behavior is also required by SGI's STL (see footnote 3). As you mentioned above, n3225 appeared without the fixes from n3066, so defect 2035 was raised; but alas the fix did not make it into the published version of C++11 (ISO/IEC 14882:2011).

此外，缺陷2035表示 a （来自 X a（r ++）; ）不能像 * a = 0 一样使用：

Furthermore, defect 2035 says that a (from X a(r++);) cannot be used like *a = 0:

在此操作之后[ie， ++ r ] r 不需要是可递增的，并且任何副本都是之前的 r 值不再需要可解除引用或可递增。

"After this operation [i.e., ++r] r is not required to be incrementable and any copies of the previous value of r are no longer required to be dereferenceable or incrementable."

在某些情况下，这可能有助于实施（就简单性而言）：参见例如在 ostream_iterator 上此问题，其中此类（无效）双增量为忽略只需返回 * this ;只有取消引用/赋值会导致 ostream_iterator 实际增加。

There are situations where this may aid the implementation (in terms of simplicity): see e.g. this question on ostream_iterator, where such (invalid) double increments are ignored simply returning *this; only a dereference/assignment causes the ostream_iterator to actually increment.

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