我要如何为int的低8位？ [英] How do i get the lower 8 bits of int?

问题描述

可以说我有一个int变量N = 8。在大多数机器上，这将是一个32位值。我怎么只能得到这样的低8位（最低字节）的二进制？此外，我怎么能访问每一个位，找出它是什么？
我的问题是有关为C

解决方案

 无符号的N = 8;
无符号low8bits = N＆安培; 0xFF的;
 

请注意以下几点：

1. 对于位操作，始终使用无符号类型


2. 位可以使用二进制遮蔽与数字提取＆安培; 运营商


3. 要访问的低8位掩码是 0xFF的，因为在二元它有它的低8位开启，其余0


4. 数字8的低8位是... 8（想想了一会儿）

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要访问大量的某一点，说 K 个位：

 无符号N = ...;
无符号kthbit =（1 <<;＆所述; k）及N;
 

现在， kthbit 将是0，如果 K 个位 N 0，有的正数（ 2 **亩）如果 K 次的位 N 1

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

解决方案

unsigned n = 8;
unsigned low8bits = n & 0xFF;

Note a few things:

1. For bitwise operations, always use the unsigned types
2. Bits can be extracted from numbers using binary masking with the & operator
3. To access the low 8 bits the mask is 0xFF because in binary it has its low 8 bits turned on and the rest 0
4. The low 8 bits of the number 8 are... 8 (think about it for a moment)

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To access a certain bit of a number, say the kth bit:

unsigned n = ...;
unsigned kthbit = (1 << k) & n;

Now, kthbit will be 0 if the kth bit of n is 0, and some positive number (2**k) if the kth bit of n is 1.

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