如何在Java中使用bitshifting [英] How to use bitshifting in Java

问题描述

我正在尝试构建一个IP头。

I am trying to construct an IP header.

IP标头包含以下字段：版本，IHL，DSCP等。我想填充一个字节数组，以便我可以以字节为单位存储信息。

An IP header has the following fields: Version, IHL, DSCP etc. I would like to populate a Byte Array such that I can store the information in bytes.

然而，我感到困惑的是版本字段只有4位宽。国际人道法也只有4位宽。如何将这两个字段的值拟合为一个字节？我需要做位移吗？

Where I get confused however is that the Version field is only 4 bits wide. IHL is also only 4 bits wide. How do I fit the values of both of those fields to be represented as a byte? Do I need to do bitshifting?

例如。版本= 4，IHL = 5.我需要创建一个等于0100 0101 = 45h或69十进制的字节。

E.g. Version = 4, IHL = 5. I would need to create a byte that would equal 0100 0101 = 45h or 69 decimal.

推荐答案

(byte) (4 << 4) | 5

这将值4移到左边，然后将低4位设置为值5. / p>

This shifts the value 4 to the left, then sets lower 4 bits to the value 5.

1. 00000100 一个值（ 4 ）


2. 01000000 向左移4位后（<<< 4 ）


3. 00000101 另一个值（ 5 ）


4. 01000101 ＃2和＃3的按位OR（ | ）的结果/ li>
   

1. 00000100 A value (4)
2. 01000000 After shifting left 4 bits (<< 4)
3. 00000101 Another value (5)
4. 01000101 The result of a bitwise OR (|) of #2 and #3

因为操作数是 int 类型（即使它们是<$ c） $ c> byte 值，当 | 等运算符时，它们会被提升为 int 对他们采取行动），最终结果需要一个演员表存储在字节。

Because the operands are int types (and even if they were byte values, they'd be promoted to int when operators like | act on them), the final result needs a cast to be stored in a byte.

如果您正在使用 byte 值作为任何按位运算中的操作数，隐式转换为 int 会导致意外结果。如果要将字节视为在该转换中未签名，请使用按位AND（& ） ：

If you are using byte values as operands in any bitwise operations, the implicit conversion to int can cause unexpected results. If you want to treat a byte as if it were unsigned in that conversion, use a bitwise AND (&):

byte b = -128; // The byte value 0x80, -128d
int uint8 = b & 0xFF; // The int value 0x00000080, 128d
int i = b; // The int value 0xFFFFFF80, -128d
int uintr = (b & 0xFF) | 0x04; // 0x00000084
int sintr = b | 0x04; // 0xFFFFFF84

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