整数溢出 - 为什么不呢 [英] Integer Overflow - Why not

问题描述

可能重复：

添加两个字符产生int

鉴于以下C ++代码：

Given the following C++ code:

unsigned char a = 200;
unsigned char b = 100;

unsigned char c = (a + b) / 2;

输出为逻辑预期的150，但是不应该有表达式中的整数溢出（a + b）？

The output is 150 as logically expected, however shouldn't there be an integer overflow in the expression (a + b)?

显然必须有一个整数提升才能处理这里溢出，或者其他一些我看不到的东西。我想知道是否有人可以启发我，所以我可以知道它是什么，不应该依赖于整数提升和溢出。

Obviously there must be an integer promotion to deal with the overflow here, or something else is happening that I cannot see. I was wondering if someone could enlighten me, so I can know what it is I can and shouldn't rely on in terms of integer promotion and overflow.

推荐答案

C ++和C都没有执行带有较小整数类型的算术计算，例如 char 和 short 。在进一步计算开始之前，这些类型几乎总是被提升为 int 。所以，你的表达式实际上被评估为

Neither C++ not C perform arithmetical computations withing "smaller" integer types like, char and short. These types almost always get promoted to int before any further computations begin. So, your expression is really evaluated as

unsigned char c = ((int) a + (int) b) / 2;

P.S。在某些奇特的平台上， int 的范围不包括 unsigned char 的范围，类型 unsigned int 将用作促销的目标类型。

P.S. On some exotic platform where the range of int does not cover the range of unsigned char, the type unsigned int will be used as target type for promotion.

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