Returning a list of strings from input

Question

I'm currently working on a function called, getBASIC(). This is why I'm making this:

Write a function getBASIC() which takes no arguments, and does the following: it should keep reading lines from input using a while loop; when it reaches the end it should return the whole program in the form of a list of strings.

The program takes input in the form of:

X GOTO Y  
Y GOTO Z  
Z END

And so on and so forth.

My code for this is as follows:

def getBASIC():
   l = []
   while len(i.split()) == 3:
      i = input()
      l.append(i)
   return(l)

Problem is, I get an UnboundLocalError: local variable 'i' referenced before assignment. Now I do know why this is, but I've suddenly become an idiot and can't figure out how to fix it. Help debugging this would be appreciated. Thanks.

I know nothing about python, but putting something like i = "A B C" before the while loop should work, I guess. — mostruash
– mostruash, Commented Mar 31, 2013 at 16:35

Benjamin · Accepted Answer · 2013-03-31 16:36:06Z

3

Simple solution

   i = input()
   l.append(i)
   while len(i.split()) == 3:
       i = input()
       l.append(i)

3 Comments

mostruash Over a year ago

First solution is simply wrong. A line is consumed and gone forever. Second one is not elegant.

Eryk Sun Over a year ago

@mostruash: the 2nd solution is a common Python idiom to keep from repeating yourself. If it's any consolation, the compiled code in 3.x doesn't actually test the boolean value of built-in True on each pass. In practice it's a do-while loop.

Dan Over a year ago

Crap. Sorry for not replying - I got sidetracked. The first solution is incorrect (I know because I already tried that.) Time to test the second one.

Dan · Accepted Answer · 2013-03-31 17:15:50Z

0

    def getBASIC():
   l = []
   x = 1
   while x == 1:
      i = input()
      l.append(i)
      if len(i.split()) != 3:
         x = 0
   return l

answered Mar 31, 2013 at 17:15

Dan

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