使用C ++中的负整数除法舍入

问题描述

假设 a 和 b 都是 int ， b 为非零。考虑在以下情况下执行 a / b 的结果：

Suppose a and b are both of type int, and b is nonzero. Consider the result of performing a/b in the following cases:

1. a 和 b 均为非负。


2. / code>和 b 均为负。


3. 其中一个为负。

1. a and b are both nonnegative.
2. a and b are both negative.
3. Exactly one of them is negative.

在情况1中，结果向下取整为最接近的整数。但标准对案例2和3有什么看法？我发现在互联网上浮动的一个旧草案表明，它是实现相关的（是的，即使情况2），但委员会倾向于使它总是向零。有谁知道（最新）标准说什么？

In Case 1 the result is rounded down to the nearest integer. But what does the standard say about Cases 2 and 3? An old draft I found floating on the Internet indicates that it is implementation dependent (yes, even case 2) but the committee is leaning toward making it always 'round toward zero.' Does anyone know what the (latest) standard says? Please answer only based on the standard, not what makes sense, or what particular compilers do.

推荐答案

根据2008年5月修订版，

According to the May 2008 revision,

您是对的：

注意75说：

C ++在这方面可能会滞后C。

Chances are that C++ will lag C in this respect. As it stands, it's undefined but they have an eye towards changing it.

我和Stroustrup和同一个委员会的成员在同一个部门工作，这是不确定的。事情需要AGES来完成，它的无休止的政治。如果它似乎愚蠢，它可能是。

I work in the same department as Stroustrup and with a member of the committee. Things take AGES to get accomplished, and its endlessly political. If it seems silly, it probably is.

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