是否定义了C ++标准算法的空白范围? [英] Is it defined to provide an empty range to C++ standard algorithms?

问题描述

接下来是我的上一个问题，我们能否证明标准允许我们将空范围传递给标准算法?

Following on from my previous question, can we prove that the standard allows us to pass an empty range to a standard algorithm?

第24.1/7段将空范围"定义为范围[i,i)(其中i是有效的)，并且i看起来像是可到达的"，但是我不确定有资格作为证明.

Paragraph 24.1/7 defines an "empty range" as the range [i,i) (where i is valid), and i would appear to be "reachable" from itself, but I'm not sure that this qualifies as a proof.

尤其是在查看排序功能时遇到麻烦.例如，std::sort:

In particular, we run into trouble when looking at the sorting functions. For example, std::sort:

复杂度:O(N log(N))(其中N == last-first)比较

Complexity: O(N log(N)) (where N == last - first) comparisons

由于通常认为log(0)是未定义的，而且我不知道0*undefined是什么，所以这里可能有问题吗?

Since log(0) is generally considered to be undefined, and I don't know what 0*undefined is, could there be a problem here?

^{(是的，好的，我有点书呆子.当然，没有自尊的 stdlib 实现不会导致将空范围传递给std::sort的实际问题.但是我想知道这里的标准措词是否有潜在的漏洞.)}

^{(Yes, ok, I'm being a bit pedantic. Of course no self-respecting stdlib implementation would cause a practical problem with an empty range passed to std::sort. But I'm wondering whether there's a potential hole in the standard wording here.)}

推荐答案

我似乎没有什么问题.在§24.1/6中，我们被告知:

I don't seem much room for question. In §24.1/6 we're told:

当且仅当存在++ i表达式的有限应用序列使i == j时，迭代器j才可从迭代器i调用.

An iterator j is called reachable from an iterator i if and only if there is a finite sequence of applications of the expression ++i that makes i == j.

以及$ 24.1/7:

and in $24.1/7:

仅当j可以从i到达时，范围[i，j)才有效.

Range [i, j) is valid if and only if j is reachable from i.

由于0是有限的，所以[i, i)是有效范围. §24.1/7继续说:

Since 0 is finite, [i, i) is a valid range. §24.1/7 goes on to say:

将库中的函数应用到无效范围的结果是 未定义.

The result of the application of functions in the library to invalid ranges is undefined.

并不能说有效范围可以保证定义的结果(合理，因为还有其他要求，例如比较功能)，但是肯定暗示范围本身是空的，不应导致UB或类似问题.但是，特别是，该标准将空范围设为另一个有效范围.空值范围和非空值范围之间没有真正的区别，因此适用于非空值有效范围的情况同样适用于空值有效范围.

That doesn't go quite so far as to say that a valid range guarantees defined results (reasonable, since there are other requirements, such as on the comparison function) but certainly seems to imply that a range being empty, in itself, should not lead to UB or anything like that. In particular, however, the standard makes an empty range just another valid range; there's no real differentiation between empty and non-empty valid ranges, so what applies to a non-empty valid range applies equally well to an empty valid range.

这篇关于是否定义了C ++标准算法的空白范围?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！