包装的4个整数作为一个字节? [英] Packing 4 Integers as ONE BYTE?
问题描述
我有四个整数{ A
, B
, C
, D
},可以有以下范围值:
I have four integers {a
, b
, c
, d
} that can have the following range of values:
A
- {0或1}(1位)
a
- {0 or 1} (1 bit)
B
- {0或1}(1位)
b
- {0 or 1} (1 bit)
C
- {0,1,2,...,7}(3位)
c
- {0, 1, 2, ..., 7} (3 bits)
D
- {0,1,2,...,7}(3位)
d
- {0, 1, 2, ..., 7} (3 bits)
首先,我想将它们包成一个字节,可以然后写入到一个二进制文件。
后来,我想解压一个字节,并从中获取到窗体的元组( A
, B
, C
, D
)。
at first, I would like to pack them into a one byte that can be then written to a binary file.
later, I would like to unpack that one byte and get from it to a tuple of the form (a
, b
, c
, d
).
我知道如何读/字节写入Python中的二进制文件。但我怎么做位打包/拆包?
I know how to read/write a byte to a binary file in python. But how do I do the bits packing/unpacking?
感谢。
推荐答案
使用移,按位或,然后转换为字符得到一个字节
Use shift and bitwise OR, then convert to a character to get a "byte":
x = chr(a | (b << 1) | (c << 2) | (d << 5))
要再次解开这个字节,先转换为整数,然后转移和使用按位与:
To unpack this byte again, first convert to an integer, then shift and use bitwise AND:
i = ord(x)
a = i & 1
b = (i >> 1) & 1
c = (i >> 2) & 7
d = (i >> 5) & 7
解释:最初,你有
Explanation: Initially, you have
0000000a
0000000b
00000ccc
00000ddd
左边的转移给你。
The left-shifts give you
0000000a
000000b0
000ccc00
ddd00000
在位或结果
dddcccba
转换为字符将其转换为一个字节。
Converting to a character will convert this to a single byte.
开箱:四个不同的右移导致
Unpacking: The four different right-shifts result in
dddcccba
0dddcccb
00dddccc
00000ddd
屏蔽(按位与)与 1
( 0b00000001
)或 7
( 0b00000111
)的结果。
0000000a
0000000b
00000ccc
00000ddd
试。
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