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nonlocal keyword in Python 2.x
I'm trying to implement a closure in Python 2.6 and I need to access a nonlocal variable but it seems like this keyword is not available in python 2.x. How should one access nonlocal variables in closures in these versions of python?
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What is this 'Lambda' everyone keeps speaking of?
... IllidanS4 wants Monica back
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answered Jul 6 '09 at 10:18
Ionuț G. StanIonuț G. Stan
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Are there any disadvantages to always using nvarchar(MAX)?
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69
Unfortunately this answer has a number of problems. It makes the 8k boundary seem like a magic number, is not true, the value gets pushed o...
Where do I find old versions of Android NDK? [closed]
...I can find older versions of the Android NDK? Our code doesn't build with r6. Surely there must be archived versions somewhere.
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Why is semicolon allowed in this python snippet?
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16
It's useful for things like timeit a = 5; a*a
– endolith
May 23 '13 at 19:00
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Why split the tag when writing it with document.write()?
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bobincebobince
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30
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What's the easiest way to escape HTML in Python?
...fe. In other words, this is not safe: <a href=" {{ html.escape(untrusted_text) }} ">
– pianoJames
Jul 30 '19 at 15:49
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Cost of len() function
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360
It's O(1) (constant time, not depending of actual length of the element - very fast) on every t...
set date in input type date
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162
Fiddle link : http://jsfiddle.net/7LXPq/93/
Two problems in this:
Date control in HTML 5 acc...
sys.argv[1] meaning in script
... prints the number of arguments, using the len function on the list.
from __future__ import print_function
import sys
print(sys.argv, len(sys.argv))
The script requires Python 2.6 or later. If you call this script print_args.py, you can invoke it with different arguments to see what happens.
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