为什么在char变量中存储255在C中赋值为-1？ [英] Why stores 255 in a char variable give its value -1 in C?

问题描述

我正在读一本C书，并且有一篇文章，作者提到：

如果ch（一个char变量）签名类型，然后在ch变量中存储255的值为-1 。

解决方案

假设8位 char s，实际上是实现定义的行为。然而，大多数实现只是存储位模式，其中对于255是



< 0xFF 。使用二进制补码解释，作为有符号的8位整数，即 -1 的位模式。在一个稀有的补码结构上，这将是负零或陷阱表示的位模式，具有符号和幅度，它将是 -127 。

如果两个假设（signedness和8位 char s）中的任何一个不成立， be¹255，因为255可以表示为一个无符号的8位整数或者是一个超过8位的有符号（或无符号）整数。

code> CHAR_BIT 至少为8，它可能会更大。

I am reading a C book, and there is a text the author mentioned:

"if ch (a char variable) is a signed type, then storing 255 in the ch variable gives it the value -1".

Can anyone elaborate on that?

解决方案

Assuming 8-bit chars, that is actually implementation-defined behaviour. The value 255 cannot be represented as a signed 8-bit integer.

However, most implementations simply store the bit-pattern, which for 255 is 0xFF. With a two's-complement interpretation, as a signed 8-bit integer, that is the bit-pattern of -1. On a rarer ones'-complement architecture, that would be the bit pattern of negative zero or a trap representation, with sign-and-magnitude, it would be -127.

If either of the two assumptions (signedness and 8-bit chars) doesn't hold, the value will be¹ 255, since 255 is representable as an unsigned 8-bit integer or as a signed (or unsigned) integer with more than 8 bits.

¹ The standard guarantees that CHAR_BIT is at least 8, it may be greater.

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