Loop Counting Scoping In Python

Question

I'm trying to figure out how this code works. How is i accessible outside the for loop?

# Palindrome of string
str=raw_input("Enter the string\n")
ln=len(str)
for i in range(ln/2) :
    if(str[ln-i-1]!=str[i]):
        break
if(i==(ln/2)-1):         ## How is i accessible outside the for loop ? 
    print "Palindrome"
else:
    print "Not Palindrome"

for loops do not have their own namespace. [Short Description of Python Scoping Rules](http://stackoverflow.com/questions/291978/short-description-of-python-scoping-rules) — Moses Koledoye, Jul 14 '16 at 17:46
`i` is accessible because loop variables stay in the current scope with their last value unless you assign something else to the name. — Morgan Thrapp, Jul 14 '16 at 17:46
Why? How or what are you implementing this for? It is overly complex for what it does. `mystr == mystr[::-1]` — Nick, Jul 14 '16 at 17:46
Good job actually asking about the specific aspect of the code you don't understand! That makes your question way more answerable. — user2357112, Jul 14 '16 at 17:48
Side note: do not use `str` as a variable name, as you are shadowing the built-in [str](https://docs.python.org/3/library/stdtypes.html#str) — idjaw, Jul 14 '16 at 17:49
Also i want understand the logic here , i am not able to get the if condition clearly , suppose my input is (nitin) so can someone guide me how its being evaluated , since the for loop will be run from i=0 to i = 2 — Kittystone, Jul 14 '16 at 17:55

score 1 · Answer 1 · edited May 23 '17 at 11:58

1

This is part of Python. Variables declared inside for loops (that includes loop counters) won't decay until they fully leave scope.

Take a look at this question:

Scoping In Python For Loops

From the answers:

for foo in xrange(10):
    bar = 2
print(foo, bar)
The above will print (9,2).

edited May 23 '17 at 11:58

Community

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answered Jul 14 '16 at 17:48

Athena

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Loop Counting Scoping In Python

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