如何使用 Python socket.send() 而不是 8 位发送一个 16 位字

Question

I trying to build up an SPI communication between my PC and a sensor-device. For that, I use a TCP/IP to SPI Converter. This side of the project is already working. I can send something over TCP/IP to the converter and it does its job.

So where is the problem? I use the socket lib. in Python 3.8 to do the TCP/IP side:

import socket, sys

TCP_IP = '192.168.100.100'
TCP_PORT = 1003
BUFFER_SIZE = 1024

s = socket.socket()

try:
    s.connect((TCP_IP, TCP_PORT))

except socket.error as mag:
    print (" Socket Erroer: " + str(mag))

s.send(b'\x8000')
data = s.recv(BUFFER_SIZE)
print(data)

What I want to do is to send (in binary ) 1000 0000 0000 0000 (= 8000 in Hex). But python split the \x8000 into \x80\x00 and send it as two 8 bit words as 1000 0000 and then a second package 0000 0000. But the converter needs it as a 16 bit word otherwise, it fills it up. And the SPI-signal is 0000 0000 1000 0000 0000 0000 0000 0000, that's terrible, I can't read a single register without the first 8 bits.

So is there a way to send /x8000 as one 16 bit word? I would also like to know more about the datatype (or whatever it is) b' something ' what is that? What does it?

Thank you for reading this. I'm open to all suggestions and critics. I hope you can help me thank you for every attempt!

Answer 1

You can use Python raw socket to include 16bits in a single packet and send.

b'something' is python notation that converts the string "something" into its binary form(bytes), by default with utf-8 .
For example if you code a=b'Hello' , a will have \x48\x65\x6C\x6C\x6F .