how to make 16bit to 8bit like that 0x0300 to 0x03 and 0x00 by python?

Question

how to make 16bit to 8bit like that 0x0300 to 0x03 and 0x00 by python?

i try to it like that

reg = 0x0300
print(reg[0:4], reg[4::])..

it show me that 'int' object is not subscriptable

reg = 0x0316
regbinary = bin(reg)
print(regbinary)
regBotoom = regbinary[-8::]
print(regBotoom)
result = hex('0b'+regBotoom)
print(result)

it show me that

0b1100010110

00010110

TypeError: 'str' object cannot be interpreted as an integer

Answer 1

You can get lower and higher 8 bits with bitwise AND and shifting:

reg = 0x0316
lowBits = reg & 0xFF
highBits = (reg >> 8) & 0xFF
print(lowBits)
print(highBits)

Answer 2

Perhaps you're after the answer @jafarlihi gave, where reg is really just an int and you shift its value to get the values you're after.

However, you may also want to see this:

import struct

reg = 790

regb = struct.pack('h', reg)
print(regb)

msb, lsb = struct.unpack('bb', regb)
print(msb, lsb)

regb_2 = struct.pack('bb', msb, lsb)
print(regb, regb_2)

value = struct.unpack('h', regb)
print(reg, value)

Result:

b'\x16\x03'
22 3
b'\x16\x03' b'\x16\x03'
790 (790,)

Answer 3

Bitwise Operators:

x = 0x4321
y = x >> 8 
  = 0x43
z = x & 0x00ff
  = 0x21

You can not manipulate an int like you are trying to do without converting it to a string first.

Answer 4

reg = 0x0300
print(hex((reg >> 8) & 0xFF))
print(hex(reg & 0xFF))

Output:

0x3
0x0