On Thu, Aug 19, 2010 at 2:01 PM, Roelof Wobben <rwob...@hotmail.com> wrote:
> <snip>
>
def is_odd(argument):
> uitkomst=is_even(argument)
> return uitkomst
>
> even=is_odd(1) ;
> if even==True :
> print "Even getal"
> if even==False:
> print "Oneven getal"
>
>
> But now I get this error message :
>
> return uitkomst
> Syntax error : return outside function.
>
Check your indention. In the email your return is at a different level than
your function. Always use 4 spaces. If your editor can't convert, you need a
better editor!
> In my opinon even calls is_odd , then uitkomst calls is_even which gives a
> true or false to uitkomst. So return uitkomst gives the outcome to even.
> But the intepreter thinks otherwise.
>
Well, good thing opinions don't count for much when talking to a computer ;)
You are correct that uitkomst contains the result of is_even, but your
return is on a different indention level. Whitespace is significant here.
If I were you, I would define is_odd this way:
def is_odd(argument):
uitkomst = not is_even(argument)
return uitkomst
> I work on a Win7 machine with Python 2.7
>
Good job providing both the traceback and your system. They make every
problem easier to debug.
HTH,
Wayne
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