问题描述
是否保证 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:
这错过了最后一句话。所以部分
Which misses the last sentence. So the part
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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