C ++运算符％保证 [英] C++ operator % guarantees

问题描述

是否保证（ - x）％m ，其中 x 和 m c> - （x％m）是正的，在c ++ 标准（c ++ 0x）

Is it guaranteed that (-x) % m, where x and m are positive in c++ standard (c++0x) is negative and equals to -(x % m)?

我知道在我所知道的所有机器上都是正确的。

I know it's right on all machines I know.

推荐答案

除了 Luchian 的答案，这是C ++ 11标准的相应部分：

In addition to Luchian's answer, this is the corresponding part from the C++11 standard:

二元的/运算符产生商，二元的％运算符
产生第一个表达式除以
秒的余数。如果/或％的第二个操作数为零，则行为是
未定义。对于积分操作数，/运算符产生代数
商，其中丢弃任何小数部分;如果商a / b是在结果类型中表示的
，（a / b）* b + a％b等于a。

The binary / operator yields the quotient, and the binary % operator yields the remainder from the division of the first expression by the second. If the second operand of / or % is zero the behavior is undefined. For integral operands the / operator yields the algebraic quotient with any fractional part discarded; if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

这错过了最后一句话。所以部分

Which misses the last sentence. So the part

（a / b）* b + a％b等于

(a/b)*b + a%b is equal to a

是唯一依赖的引用，这意味着 a％b 将始终具有 a ，因为 / 的截断行为。所以如果你的实现在这方面遵守C ++ 11标准，模运算的符号和值的确完美地定义为负操作数。

Is the only reference to rely on, and that implies that a % b will always have the sign of a, given the truncating behaviour of /. So if your implementation adheres to the C++11 standard in this regard, the sign and value of a modulo operation is indeed perfectly defined for negative operands.

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