C语言 我如何获得int的低8位？
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How do i get the lower 8 bits of int?
提问by Laz
Lets say I have an int variable n = 8. On most machines this will be a 32 bit value. How can I only get the lower 8 bits (lowest byte) of this in binary? Also how can i access each bit to find out what it is? My question is related to C
假设我有一个 int 变量 n = 8。在大多数机器上，这将是一个 32 位值。我怎样才能以二进制形式获得它的低 8 位（最低字节）？另外我如何访问每一位以找出它是什么？我的问题与C有关
回答by Eli Bendersky
unsigned n = 8;
unsigned low8bits = n & 0xFF;
Note a few things:
请注意以下几点：
- For bitwise operations, always use the
unsignedtypes - Bits can be extracted from numbers using binary masking with the
&operator - To access the low 8 bits the mask is
0xFFbecause in binary it has its low 8 bits turned on and the rest 0 - The low 8 bits of the number 8 are... 8 (think about it for a moment)
- 对于按位运算，始终使用
unsigned类型 - 可以使用带有
&运算符的二进制掩码从数字中提取位 - 要访问低 8 位，掩码是
0xFF因为在二进制中，它的低 8 位打开，其余为 0 - 数字8的低8位是... 8（想一想）
To access a certain bit of a number, say the kth bit:
要访问数字的某个位，请说第kth 位：
unsigned n = ...;
unsigned kthbit = (1 << k) & n;
Now, kthbitwill be 0 if the kth bit of nis 0, and some positive number (2**k) if the kth bit of nis 1.
现在，kthbit如果k第0位n为 0，则为 0，2**k如果第k1 位n为 1 ，则为某个正数 ( ) 。
回答by Mark Rushakoff
Use bitwise arithmetic to mask off the lowest 8 bits:
使用按位算术屏蔽掉最低的 8 位：
unsigned char c = (x & 0xFF);
To access the nth lowest bit, the equation is (x & (1 << n))(n of zero indicates the least significant bit). A result of zero indicates the bit is clear, and non-zero indicates the bit is set.
要访问第 n 个最低位，等式是(x & (1 << n))（n 为零表示最低有效位）。结果为零表示该位已清除，非零表示该位已设置。
回答by njsf
The best way is to use the bit logical operator & with the proper value.
最好的方法是使用具有适当值的位逻辑运算符 &。
So for the lower 8 bits:
所以对于低 8 位：
n & 0xFF; /* 0xFF == all the lower 8 bits set */
Or as a general rule:
或者作为一般规则：
n & ((1<<8)-1) /* generate 0x100 then subtract 1, thus 0xFF */
You can combine with the bit shift operator to get a specific bit:
您可以结合位移运算符来获取特定位：
(n & (1<<3))>>3;
/* will give the value of the 3rd bit - note the >>3 is just to make the value either 0, or 1, not 0 or non-0 */
回答by user11977
You can test if a particular bit is set in a number using << and &, ie:
您可以使用 << 和 & 测试是否在数字中设置了特定位，即：
if (num & (1<<3)) ...
if (num & (1<<3)) ...
will test if the fourth bit is set or not.
将测试是否设置了第四位。
Similarly, you can extract just the lowest 8 bits (as an integer) by using & with a number which only has the lowest 8 bits set, ie num & 255or num & 0xFF(in hexadecimal).
类似地，您可以通过将 & 与一个仅设置了最低 8 位的数字一起使用来提取最低 8 位（作为整数），即num & 255或num & 0xFF（以十六进制表示）。
