Returning variable outside function

Question

Why is x not defined outside the function? Is return x placed wrong?

def find():

x else:

find()

You are breaking before you are returning, and should your conditional statement evaluate to false you aren't returning at all. — miradulo, Apr 23 '15 at 14:25
`break` immediately exits the loop, so the statement after it is never executed. — Barmar, Apr 23 '15 at 14:25
please review http://stackoverflow.com/questions/291978/short-description-of-python-scoping-rules — Dwight J. Browne, Apr 23 '15 at 14:26

score 6 · Answer 1 · answered Apr 23 '15 at 14:27

Put return x outside the loop:

def find(): 
    data=file('file.dat')
    x=0
    for line in data:
        if 'metaend' in line:
            break
        else:
            x+=1
    return x

To get the value of x, you use the return value of the function:

count = find()

score 2 · Answer 2 · edited Apr 23 '15 at 14:30

Your function never returns anything. Check this, with some added error handling for no end of metadata detection

def find(): 
    data=file('file.dat')
    x=0
    for line in data:
        if 'metaend' in line:
            return x

        x += 1
    raise Exception('heeey no end of metadata')

By the way, python has a very nice function for a counter in a loop:

def find(): 
    data=file('file.dat')

    for counter, line in enumerate(data):
        if 'metaend' in line:
            return counter

    raise Exception('heeey no end of metadata')

Returning variable outside function

2 Answers2