Any hex str of value above \x7F causes an extra byte to printed.

python3 -c "print('\x7F')" > test.txt && xxd test.txt
00000000: 7f0a                                     ..                               

python3 -c "print('\x80')" > test.txt && xxd test.txt
00000000: c280 0a                                  ...

If you're trying to get raw bytes, you need to use

    print(b'\x80')

what's happening right now is that the '\x80' is treated as a unicode code point (see https://docs.python.org/3/howto/unicode.html#the-string-type), and when Python goes to print it, it gets encoded to the raw underlying bytes. Which, in the default encoding of utf-8 requires the extra byte.

>>> '\x80'.encode()
b'\xc2\x80'

However print() called with non-str argument would firstly call str() on it, which is most likely not what reporter wanted:

>>> print(b'\x80')
b'\x80'
>>> str(b"\x80")
"b'\\x80'"
>>> print(str(b"\x80"))
b'\x80'

$ python -c "print(b'\x80')" > test.txt
$ xxd test.txt
00000000: 6227 5c78 3830 270a                      b'\x80'.


Proper solution is to write to files opened in binary mode, which in case of stdout and stderr means to use sys.stdout.buffer and sys.stderr.buffer:

>>> sys.stdout.buffer.write(b"\x80")
�1

$ python -c "import sys; sys.stdout.buffer.write(b'\x80')" > test.txt
$ xxd test.txt
00000000: 80                                       .

Thanks, Arfrever

Seems unnecessarily complicated for what worked in Python2.

@Artificial please see the various blog posts and explanations about why the clear separation between bytes and text came to be and thus this change isn't "unnecessarily complicated".