当 sizeof(char) != 1 时 C 中的字节精度指针算术 [英] Byte precision pointer arithmetic in C when sizeof(char) != 1

问题描述

如何以单字节精度便携地执行指针运算?

How can one portably perform pointer arithmetic with single byte precision?

请记住:

• char 在所有平台上都不是 1 字节
• sizeof(void) == 1 仅在 GCC 中作为扩展提供
• 虽然某些平台可能有指针 deref 指针对齐限制，但算术可能仍需要比最小基本 POD 类型的大小更精细的粒度

• char is not 1 byte on all platforms
• sizeof(void) == 1 is only available as an extension in GCC
• While some platforms may have pointer deref pointer alignment restrictions, arithmetic may still require a finer granularity than the size of the smallest fundamental POD type

推荐答案

您的假设是有缺陷的 - sizeof(char) 被定义 为 1.

Your assumption is flawed - sizeof(char) is defined to be 1 everywhere.

来自 C99 标准 (TC3)，在第 6.5.3.4 节(sizeof 运算符")中:

From the C99 standard (TC3), in section 6.5.3.4 ("The sizeof operator"):

(第 2 段)

sizeof 运算符产生大小(以字节为单位)其操作数，这可能是一个表达式或带括号的类型名称.

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type.

(第 3 段)

当应用于具有输入 char、unsigned char 或 signedchar，(或合格的版本其中)结果为 1.

When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1.

当将这些放在一起时，很明显在 C 中，无论 char 是什么大小，该大小都是一个字节"(即使在某些给定平台上超过 8 位).

When these are taken together, it becomes clear that in C, whatever size a char is, that size is a "byte" (even if that's more than 8 bits, on some given platform).

char 因此是最小的可寻址类型.如果您需要以小于 char 的单位寻址，您唯一的选择是一次读取 char 并使用位运算符来屏蔽 的部分>char 你想要的.

A char is therefore the smallest addressable type. If you need to address in units smaller than a char, your only choice is to read a char at a time and use bitwise operators to mask out the parts of the char that you want.

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