Python，如何解码二进制编码的十进制(BCD)

2022-01-09 00:00:00 python binary decode struct bcd

问题描述

二进制字段的描述是:

来电号码，用压缩的BCD码表示，多余的位用0xF"填充

Caller number, expressed with compressed BCD code, and the surplus bits are filled with "0xF"

我尝试使用结构格式 '16c' 进行打印，我得到:('3', 'x00', 'x02', 'x05', 'x15'、'x13'、'G'、'O'、'xff'、'xff'、'xff'、'xff'、'xff'、'xff'、'xff', 'xff') 如果我使用 '16b' 我得到 (51, 0, 2, 5, 21, 19, 71, 79, -1, -1、-1、-1、-1、-1、-1、-1).而且不正确，我应该得到电话号码，上面的号码是无效的.

I have tried to print with struct format '16c' and I get: ('3', 'x00', 'x02', 'x05', 'x15', 'x13', 'G', 'O', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff') and if I use '16b' i get (51, 0, 2, 5, 21, 19, 71, 79, -1, -1, -1, -1, -1, -1, -1, -1). And it is not correct, I should get phone number, and numbers above are invalid.

print struct.unpack_from('>16b', str(data.read()),offset=46)

上面是无效的代码，我得到了无效的数字.我应该用什么格式解压那个 16 字节的字段以及如何转换 BCD 码?

Above is code that didn't work and I get invalid numbers. With what format should I unpack that 16 byte field and how to convert BCD code ?

解决方案

BCD 码使用每个数字 4 位，通常只编码数字 0 - 9.所以序列中的每个字节包含 2 个数字，每 4 位 1 个信息.

BCD codes work with 4 bits per number, and normally encode only the digits 0 - 9. So each byte in your sequence contains 2 numbers, 1 per 4 bits of information.

以下方法使用生成器来生成这些数字；我假设 0xF 值意味着没有更多的数字可以跟随:

The following method uses a generator to produce those digits; I am assuming that a 0xF value means there are no more digits to follow:

def bcdDigits(chars):
    for char in chars:
        char = ord(char)
        for val in (char >> 4, char & 0xF):
            if val == 0xF:
                return
            yield val

这里我使用 右移运算符 向左移动- 最右边 4 位，和一个 按位与 来选择最右边的 4 位.

Here I use a right-shift operator to move the left-most 4 bits to the right, and a bitwise AND to select just the right-most 4 bits.

演示:

>>> characters = ('3', 'x00', 'x02', 'x05', 'x15', 'x13', 'G', 'O', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff')
>>> list(bcdDigits(characters))
[3, 3, 0, 0, 0, 2, 0, 5, 1, 5, 1, 3, 4, 7, 4]

该方法适用于 c 输出；如果直接传递整数，则可以跳过方法中的 ord 调用(但请改用 B 无符号变体).或者，您可以直接从文件中读取这 16 个字节，然后直接将此函数应用于这些字节，而无需使用 struct.

The method works with the c output; you can skip the ord call in the method if you pass integers directly (but use the B unsigned variant instead). Alternatively, you could just read those 16 bytes straight from your file and apply this function to those bytes directly without using struct.

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