remove is-even check from Code128 look_next()

Okay so I'm not totally sure I understand the logic behind the is-even
check here. It was added as part of a commit intended to optimise the
barcode width.

So, from what I can see, there are two rules for whether to switch to 128C:

- If the 128C section is followed by another charset switch, it should
  be at least 5 digits long to save space.
- If it is the last charset of the barcode, it should be at least 4
  digits long to save space.

As long as these two things are true, you will always see a space saving
over not switching.

Now, the code only checks if it's at least 4 digits, but that's fine
since it's simpler and just means that there is a case where we will
switch to 128C without saving any width, but also without incurring any
additional width.

Correct me if I'm wrong, but I don't think evenness comes into it - a
single digit encoded in 128C does not itself save space, but may as well
be included in the rest of the 128C block which we already know will
save space because it's at least 4 characters long.

Because of the evenness check, there's a case where we can create a
longer barcode. When the barcode starts with a sequence of 5 digits, the
first digit will be encoded in 128B because of the odd number of digits,
incurring an additional charset switch and widening the barcode.

I'm wondering if anyone has an example of a barcode which is shorter
because of the evenness check. Otherwise I think we should go ahead and
remove it as proposed.

Author: audoh-tickitto

barcode/codex.py

@@ -169,7 +169,7 @@ def look_next():
                     digits += 1
                 else:
                     break
-            return digits > 3 and (digits % 2) == 0
+            return digits > 3
 
         codes = []
         if self._charset == "C" and not char.isdigit():