我怎样才能访问特定比特组从一个变量？ [英] How can I access a specific group of bits from a variable?
问题描述
我有位的X号的变量。我怎样才能提取特定比特组，然后在C对他们的工作？
I have a variable with "x" number of bits. How can I extract a specific group of bits and then work on them in C?
推荐答案
您将与一系列的2位逻辑运算的做到这一点。
You would do this with a series of 2 bitwise logical operations.
[[术语的MSB（最高有效位）是最-显著位; LSB（LSB）是最不显著位。假设位从LSB == 0到一些MSB（例如在32位机器上31）。比特位置i重新$ P $的值psents整数的2 ^ I分量的系数。]]
[[Terminology MSB (msb) is the most-significant-bit; LSB (lsb) is the least-significant-bit. Assume bits are numbered from lsb==0 to some msb (e.g. 31 on a 32-bit machine). The value of the bit position i represents the coefficient of the 2^i component of the integer.]]
例如，如果您有 INT X ，并要提取比特×[msb..lsb]包容性的，例如一个4位字段x的一定范围内[7..4]出X [31..0]位，那么：
For example if you have int x, and you want to extract some range of bits x[msb..lsb] inclusive, for example a 4-bit field x[7..4] out of the x[31..0] bits, then:
-
通过由LSB位，如移动X右
X＆GT;＆GT; LSB ，你把x的LSB位在第0除权pression的（至少显著）位，这也正是它需要的。
By shifting x right by lsb bits, e.g.
x >> lsb, you put the lsb bit of x in the 0th (least significant) bit of the expression, which is where it needs to be.
现在，你必须屏蔽掉上面的那些MSB指定的任何剩余位。这样的位数是MSB LSB + 1，我们可以形成的1位位掩码字符串长与前pression 〜（〜0℃;＆LT;（MSB- LSB + 1））。例如〜（〜0℃;＆LT;（7-4 + 1））==〜0b11111111111111111111111111110000 == 0b1111
Now you have to mask off any remaining bits above those designated by msb. The number of such bits is msb-lsb + 1. We can form a bit mask string of '1' bits that long with the expression ~(~0 << (msb-lsb+1)). For example ~(~0 << (7-4+1)) == ~0b11111111111111111111111111110000 == 0b1111.
全部放在一起，你可以提取要与位向量与这个前pression一个新的整数：
Putting it all together, you can extract the bit vector you want with into a new integer with this expression:
(x >> lsb) & ~(~0 << (msb-lsb+1))
例如，
int x = 0x89ABCDEF;
int msb = 7;
int lsb = 4;
int result = (x >> lsb) & ~(~0 << (msb-lsb+1));
// == 0x89ABCDE & 0xF
// == 0xE (which is x[7..4])
请有意义吗？
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