Detecting empty string when decoding byte array into unicode? (Python)

Question

I'm trying to read an array of bytes character by character and decode it to a unicode string, as below:

current_character = byte_array[0:1].decode("utf-8")

For each character, I'm trying to check whether the result of .decode("utf-8") equals the empty string, but I can't seem to be able to detect this. When I print out the result of decoding, I get the empty string. But how do I translate this detection into code?

I've tried:

if not current_character

if current_character is u""

But they both don't work. Any suggestions?

How about current_character = byte_array.decode("utf-8")[0:1]? — Puffin GDI
– Puffin GDI, Commented Nov 10, 2014 at 5:35
What do you mean by they don't work? Also, can you add a sample value for byte_array? — Anshul Goyal
– Anshul Goyal, Commented Nov 10, 2014 at 5:57
unrelated: utf-8 is a variable-length encoding: it is incorrect to decode a single byte at a time if arbitrary utf-8 encoded input is allowed. — jfs
– jfs, Commented Apr 29, 2016 at 15:31

regapictures · Accepted Answer · 2016-04-28 20:40:34Z

5

Binary strings seem to be written like b'some data', so try this:

if current_character == b'':
    # Code

answered Apr 28, 2016 at 20:40

regapictures

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Comments

jfs · Accepted Answer · 2016-04-29 15:30:24Z

1

if not current_character: is the correct way to find out whether current_character is an empty string in Python. It is equivalent to if len(current_character) == 0:

>>> not ''
True
>>> not b''
True

If it doesn't work for you then something else is broken in your code.

answered Apr 29, 2016 at 15:30

jfs

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Comments

Alfabravo · Accepted Answer · 2014-11-10 05:58:25Z

0

try this:

if current_character == '':
    print('empty string')

edited Nov 10, 2014 at 5:58

Alfabravo

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answered Nov 10, 2014 at 5:38

NeoWu

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Artem Sokolov · Accepted Answer · 2014-11-10 06:33:04Z

0

No need to decode one byte, just check if it equals 0

current_character = byte_array[x:x+1]
if current_character == 0x00:
    pass #...

answered Nov 10, 2014 at 6:33

Artem Sokolov

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