Detailed analysis of message signals occupying multiple bytes and bits

In the actual application message parsing program, it is necessary to account for 20 Bit for data analysis , Because I haven't written the program here for message parsing for a long time, I still have a good time , The calculation process is now recorded for your reference , It can be used when a message signal occupying multiple bytes is encountered during subsequent parsing .

Received message signal （20 position , Take two half bytes ）：

1010 10110011 10111011    

As shown above ： Red is the first byte , Lower eight ; Blue is the second byte , Middle eight ; Green is the third byte , high 4 position .

1010 10110011 10111011   Convert decimal to zero    701371

10110011 10111011   Convert decimal to zero    46011

10111011   Convert decimal to zero    187

（10110011（ Convert decimal to zero 179）） * 256 = 45824

46011 - 187 = 45824 It can be concluded that the middle eight bits are binary bit multiplication 256（2 The octave of ） Work out ;

After verification, the same principle can be obtained The upper four bits are the binary bits multiplied by 256*256（2 To the sixteenth power of ） Work out ;

if 20 The lowest digit 4 Bit is the upper four bits of the first byte filled , Move right 4 Bit to put it at the lowest 4 position , Last pass &0x0F Work out the result

The parsing process is that the lowest four digits are directly converted to decimal , The middle eight digits are converted to decimal and multiplied by 16（ That is to say 2 Of 4 Power ）, The high octave is converted to decimal and multiplied by 16*256（ That is to say 2 Of 12 Power ）, Add to get the final decimal result .

     Of course , If the number of digits is small , There are more ways to think of , By shifting 、 Bit or 、 And / or cleverly disposed of , In addition, the binary data string of the message is directly converted to decimal number after being intercepted by the function , Another way of thinking is a good way .