Python Source Code encodings does not work

Question

On my Paspberry I run a Python script under Python 2.7.

I want to declare some string literals as Latin-1, not as UTF-8.

Therefore I added the statement

# -*- coding: latin-1 -*-

at start of my file. But regardless of which coding I'm use the following code snippet always declares my string as UTF-8.

    s = 'äöü'
    print '%s %d' %(s, len(s))
    print '%x %x %x %x %x %x' % (ord(s[0]), ord(s[1]), ord(s[2]), ord(s[3]), ord(s[4]), ord(s[5]))

Shows me always:

äöü 6
c3 a4 c3 b6 c3 bc

What is the correct way to declare a string literal with Latin-1 coding, i.e. In my case I would expect a string with 3 characters: 0xe4, 0xf6, 0xfc?

Community · Accepted Answer · 2017-05-23 12:31:05Z

2

If I understand, you can use:

s.encode('latin-1');

for your problem.

Example:

>>> s = u'ééé'.encode('latin1') 
>>> s.decode('latin1')
u'\xe9\xe9\xe9'

Source of the example.

Give me feedback if you can do it.

edited May 23, 2017 at 12:31

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answered Feb 11, 2016 at 22:31

PRVS

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3 Comments

AndiR Over a year ago

yes that works bot I loked for a simpler solution to specify some plain string literals.

PRVS Over a year ago

How simpler? Be more specific, so I can help you, maybe.

AndiR Over a year ago

OK, maybe I don't understand python's philosophy. In other (C-based) languages I declare char s[] = 'äöü' and I get a string with exact 3 characters. Python seems to do some 'magic' around this; but maybe it's a feature I don't understand. But for which the coding comment at start of file is necessary than? It doesn't change anything regardless what I specify - utf-8 or latin-1.

Mike Samuel · Accepted Answer · 2016-03-05 05:06:29Z

0

'\xe4\xf6\xfc'

is a byte string with the 3 bytes you specified.

As to why it's not encoding properly, that depends on how your editor is saving the file. You might want to check that it's being saved using Latin-1 by looking at your source file with hexdump.

answered Mar 5, 2016 at 5:06

Mike Samuel

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