从 4 字节十六进制数中读取经度 [28:0]

Question

I am receiving a longitude and accuracy as a 4 byte hexadecimal string: 99054840 I'm trying to extract a longitude from this value.

The specs tell me the following:

• Bits [28:0]: signed value λ, little-endian format, longitude in ° = λ ÷ 1,000,000

• Bits [31:29]: unsigned value α, range 0-7, a measure for accuracy

My device is located physically at a longitude of 4.7199. So I now what the result of the conversion should be.

To read the value of the longitude I currently do (with incorrect result):

def get_longitude(reading):
    # split in different bytes
    n=2
    all_bytes = [reading[i:i+n] for i in range(0, len(reading), n)]
    
    # convert to binary
    long_bytes_binary = list(map(hex_to_binary, all_bytes))
    
    # drop the accuracy bits
    long_bytes_binary[3] = long_bytes_binary[3][0:5]

    # little endian & concatenate bytes
    longitude_binary = ''.join(list(reversed(long_bytes_binary)))

    # get longitude
    lon = binary_to_decimal(int(longitude_binary))/1_000_000

Which comes to 138.93. So totally different from the 4.7199 (expected outcome)

Here are the helper methods

def hex_to_binary(payload):
    scale = 16
    num_of_bits = 8
    binary_payload = bin(int(payload, scale))[2:].zfill(num_of_bits)
    return binary_payload   

def binary_to_decimal(binary): 
  binary1 = binary
  decimal, i, n = 0, 0, 0
  while(binary != 0):
    dec = binary % 10
    decimal = decimal + dec * pow(2, i)
    binary = binary//10
    i += 1
  return decimal 

What am I doing wrong? How can I correctly read the value? Or is my device broken:)

Answer 1

The OP code dropped the last 3 bits instead of the first 3 bits for accuracy. This change fixes it:

# drop the accuracy bits
long_bytes_binary[3] = long_bytes_binary[3][3:]

But the calculation can be much more simple:

def hex_to_longitude(x):
    b = bytes.fromhex(x)             # convert hex string to bytes
    i = int.from_bytes(b,'little')   # treat bytes as little-endian integer
    return (i & 0x1FFFFFFF) / 1e6    # 29-bitwise AND mask divided by one million

x = '99054840'
print(hex_to_longitude(x))

4.720025

Answer 2

I'm cheating a little bit here by using struct to do the endian swap, but you get the idea.

import struct

val = 0x99054840
val = struct.unpack('<I',struct.pack('>I',val))[0]
print(hex(val))
accuracy = (val >> 29) & 7
longitude = (val & 0x1ffffff) / 1000000
print(accuracy,longitude)

Output:

C:\tmp>x.py
0x40480599
2 4.720025

C:\tmp>                                               ```