这是什么意思按位左移16个无符号的字符 [英] what does it mean to bitwise left shift an unsigned char with 16

问题描述

我读包含 .cpp文件unsigned char型变量，它试图按位左移16位，因为一个 unsigned char型由8位左移16位将清除所有位和八个0填充它。

i am reading a .cpp file containing a unsigned char variable, it's trying the bitwise left shift 16 bits, since an unsigned char is composed of 8 bits, left shift 16 bits will erase all the bits and fill it with eight 0s.

无符号字符字节= 0xFF的;字节＆LT;＆LT; 16;

推荐答案

当你转移的值，

unsigned char x = ...;
int y = x << 16;

类型的X 如果晋升为 INT unsigned char型在 INT （大多数系统）配合，或无符号如果 unsigned char型不适合在 INT （罕见 ¹ ）。只要你的 INT 宽或更宽25位，则没有数据将被丢弃 ² 。

The type of x is promoted to int if unsigned char fits in an int (most systems), or to unsigned if unsigned char does not fit in an int (rare¹). As long as your int is 25 bits wide or wider, then no data will be discarded².

请注意，这是完全无关的事实，即 16 的类型为 INT 。

Note that this is completely unrelated to the fact that 16 has type int.

/* All three are exactly equivalent */
x << 16;
x << 16u;
x << (unsigned char) 16;

来源：从n1516（C99草案）：

Source: from n1516 (C99 draft):

§6.5.7第3款：位运算移位操作符

§6.5.7 paragraph 3: Bitwise Shift Operators

整数促销活动在每个操作数执行。的结果的类型是
  该促进左操作数。

The integer promotions are performed on each of the operands. The type of the result is that of the promoted left operand.

§6.3.1.1第2款规定：布尔，字符和整数

§6.3.1.1 paragraph 2: Boolean, characters, and integers

如果int可以重新present原始类型的所有值（由宽度的限制，对于
  位字段），值转换为int;否则，它被转换为一个无符号
  int类型。这些被称为整数促销。

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.

脚注：

¹ ：有些DSP芯片以及某些Cray超级已知有的sizeof（字符）==的sizeof（INT）。这简化了处理器的加载存储单元设计在额外的内存消耗的成本。

¹: Some DSP chips as well as certain Cray supercomputers are known to have sizeof(char) == sizeof(int). This simplifies design of the processor's load-store unit at the cost of additional memory consumption.

² ：如果您的左移晋升为 INT ，然后溢出 INT ，这是不确定的行为（魔鬼可能飞出你的鼻子）。相比之下，满溢的无符号总是被明确定义，所以移位应的通常可在无符号类型。

²: If your left shift is promoted to int and then overflows the int, this is undefined behavior (demons may fly out your nose). By comparison, overflowing an unsigned is always well-defined, so bit shifts should usually be done on unsigned types.

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