增量运算符在char类型的边界处的行为 [英] Behavior of increment operator at bounds for char type

问题描述

我想知道C ++在这种情况下的行为：

I wonder how C++ behaves in this case:

char variable = 127;
variable++;

在这种情况下，变量现在等于-128。

In this case, variable now equals to -128. However did the increment operator wrapped the value to its lower bound or did an overflow occurred?

提前感谢！

推荐答案

溢出发生并导致未定义行为。

Section 5.5:

如果在表达式的计算中，结果不是
数学定义的，或不在可表示值的范围内

Ifduring the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined [...]

标准注意到整数溢出是，在大多数实现中，被忽略。但这并不代表保证。

The standard goes on to note that integer overflows are, in most implementations, ignored. But this doesn't represent a guarantee.

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