could someone explain to me why some of the integers passed to to_bytes() method gives a strange result?
>>> b = 5152
>>> b.to_bytes(2, byteorder='big')
b'\x14 '
>>> b = 5153
>>> b.to_bytes(2, byteorder='big')
b'\x14!'
>>> b=16592
>>> b.to_bytes(2, byteorder='big')
b'@\xd0'
How to interpret '@' '!' ' '? For 16592 I expected b'\\x40\\xd0'.
I read Python 3 documentation and all examples from there work fine. Python 3 to_byte() description .
>>> b=1024
>>> b.to_bytes(2, byteorder='big')
b'\x04\x00'
I also try an example for this Stackoverflow post and it works like a charm.
>>> b = 1245427
>>> b.to_bytes(3, byteorder='big')
b'\x13\x00\xf3'
Aditional info:
Python 3.6.4 (default, Feb 1 2018, 11:06:09)
[GCC 7.2.1 20170915 (Red Hat 7.2.1-2)] on linux
Program runs on Fedora 27.
If the value of a byte, interpreted as ASCII, is printable, then the repr()
of that byte is the printable character.
Since the ASCII value of @
is 0x40, these two values are equivalent b'@'
, b'\\x40'
.
This may be more easily seen through demonstration than an explanation:
>>> b'\x40'
b'@'
But regardless of representation, that object is a bytes
of length 1, with a value of 64 in the first byte:
>>> b'\x40'[0] == 64
True
>>> b'@'[0] == 64
True
Returning to your example, if you want to know the hex value of each byte, you can use bytes.hex()
:
>>> b=16592
>>> b.to_bytes(2, byteorder='big').hex()
'40d0'
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