C++ 从一个字节中抓取 n 位
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Grabbing n bits from a byte
提问by user1871869
I'm having a little trouble grabbing n bits from a byte.
我在从一个字节中获取 n 位时遇到了一些麻烦。
I have an unsigned integer. Let's say our number in hex is 0x2A, which is 42 in decimal. In binary it looks like this: 0010 1010. How would I grab the first 5 bits which are 00101 and the next 3 bits which are 010, and place them into separate integers?
我有一个无符号整数。假设我们的十六进制数是 0x2A，十进制数是 42。在二进制中它看起来像这样：0010 1010。我如何获取前 5 位 00101 和接下来的 3 位 010，并将它们放入单独的整数中？
If anyone could help me that would be great! I know how to extract from one byte which is to simply do
如果有人可以帮助我，那就太好了！我知道如何从一个字节中提取，这是简单的
int x = (number >> (8*n)) & 0xff // n being the # byte
which I saw on another post on stack overflow, but I wasn't sure on how to get separate bits out of the byte. If anyone could help me out, that'd be great! Thanks!
我在另一篇关于堆栈溢出的帖子中看到过，但我不确定如何从字节中获取单独的位。如果有人可以帮助我，那就太好了！谢谢！
回答by rici
Integers are represented inside a machine as a sequence of bits; fortunately for us humans, programming languages provide a mechanism to show us these numbers in decimal (or hexadecimal), but that does not alter their internal representation.
整数在机器内部表示为位序列；对我们人类来说幸运的是，编程语言提供了一种机制来向我们显示这些数字的十进制（或十六进制），但这不会改变它们的内部表示。
You should revise the bitwise operators &, |, ^and ~as well as the shift operators <<and >>, which will help you understand how to solve problems like this.
您应该修改位运算符&，|，^和~以及移位运算符<<和>>，这将有助于您了解如何解决这类问题。
The last 3 bits of the integer are:
整数的最后 3 位是：
x & 0x7
The five bits starting from the eight-last bit are:
从最后八位开始的五位是：
x >> 3 // all but the last three bits
& 0x1F // the last five bits.
回答by Mats Petersson
"grabbing" parts of an integer type in C works like this:
在 C 中“抓取”整数类型的部分是这样的：
- You shift the bits you want to the lowest position.
- You use
&to mask the bits you want - ones means "copy this bit", zeros mean "ignore"
- 您将想要的位移到最低位置。
- 你
&用来屏蔽你想要的位 - 一个表示“复制这个位”，零表示“忽略”
So, in you example. Let's say we have a number int x = 42;
所以，在你的例子中。假设我们有一个数字int x = 42;
first 5 bits:
前 5 位：
(x >> 3) & ((1 << 5)-1);
or
或者
(x >> 3) & 31;
To fetch the lower three bits:
获取低三位：
(x >> 0) & ((1 << 3)-1)
or:
或者：
x & 7;
回答by gaborsch
Say you want hibits from the top, and lobits from the bottom. (5 and 3 in your example)
假设您想要hi顶部的lo位，以及底部的位。（在您的示例中为 5 和 3）
top = (n >> lo) & ((1 << hi) - 1)
bottom = n & ((1 << lo) - 1)
Explanation:
解释：
For the top, first get rid of the lower bits (shift right), then mask the remaining with an "all ones" mask (if you have a binary number like 0010000, subtracting one results 0001111- the same number of 1s as you had 0-s in the original number).
对于顶部，首先去掉低位（右移），然后用“全为”掩码掩码剩余的位（如果您有一个二进制数，例如0010000，减去一个结果0001111- 与1您拥有的0s数量相同-s在原来的号码）。
For the bottom it's the same, just don't have to care with the initial shifting.
对于底部，它是相同的，只是不必关心初始移位。
top = (42 >> 3) & ((1 << 5) - 1) = 5 & (32 - 1) = 5 = 00101b
bottom = 42 & ((1 << 3) - 1) = 42 & (8 - 1) = 2 = 010b
回答by wirrbel
You could use bitfields for this. Bitfields are special structs where you can specify variables in bits.
您可以为此使用位域。位域是特殊的结构，您可以在其中以位为单位指定变量。
typedef struct {
unsigned char a:5;
unsigned char b:3;
} my_bit_t;
unsigned char c = 0x42;
my_bit_t * n = &c;
int first = n->a;
int sec = n->b;
Bit fields are described in more detail at http://www.cs.cf.ac.uk/Dave/C/node13.html#SECTION001320000000000000000
位字段在http://www.cs.cf.ac.uk/Dave/C/node13.html#SECTION001320000000000000000中有更详细的描述
The charm of bit fields is, that you do not have to deal with shift operators etc. The notation is quite easy. As always with manipulating bits there is a portability issue.
位域的魅力在于，您不必处理移位运算符等。表示法非常简单。与操作位一样，存在可移植性问题。
回答by meyumer
int x = (number >> 3) & 0x1f;
int x = (number >> 3) & 0x1f;
will give you an integer where the last 5 bits are the 8-4 bits of numberand zeros in the other bits.
会给你一个整数，其中最后 5 位是 8-4 位，number其他位为零。
Similarly,
相似地，
int y = number & 0x7;
int y = number & 0x7;
will give you an integer with the last 3 bits set the last 3 bits of numberand the zeros in the rest.
会给你一个整数，最后 3 位设置最后 3 位，number其余的为零。
回答by Dave
just get rid of the 8* in your code.
去掉代码中的 8* 即可。
int input = 42;
int high3 = input >> 5;
int low5 = input & (32 - 1); // 32 = 2^5
bool isBit3On = input & 4; // 4 = 2^(3-1)
