Python: Using a variable inside a function (but the variable should be users input)

Question

I am trying to write a little hangman programme and I have to use this function in it (it is explicitly said in the assignment). It should check whether the guessed letter is in the word and if so, return a word consisting of *'s and guessed letter on the right position. Yet I don`t understand how can I use the "letter" variable if it is the one the user types in (his guess), so it is practically assigned outside the function. I am stuck a bit and I would be grateful if someone would say how to do it.

def print_guessed():
    global secret_word
    global letters_guessed
    word = []
    for i in range (0, len(secret_word)):
        if letter == secret_word[i]: word.append(letter)
        else: word.append('-')
    word_str = ''.join(word)
    return word_str

My thoughts exactly. If you have to use this function and are not allowed to modify it, you're screwed. It will throw at least a NameError because letter isn't defined anywhere. — Tim Pietzcker
– Tim Pietzcker, Commented Apr 5, 2013 at 20:17
Don't see why you need to globalise stuff; use arguments in your function. — TerryA
– TerryA, Commented Apr 5, 2013 at 20:18
@Pyrce: This is horrid! Hey...wait a minute...are you the prof? — Gerrat
– Gerrat, Commented Apr 5, 2013 at 20:18

Aswin Murugesh · Accepted Answer · 2013-04-05 20:20:15Z

2

I f you can make change to the function, then modify an argument to it, which is the variable-letter.

def print_guessed(letter):
    #do_stuff

And supply the letter guessed in the call statement

print_guessed(letter)

answered Apr 5, 2013 at 20:20

Aswin Murugesh

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5 Comments

DaveTheScientist Over a year ago

Yup, I was just going to write that. Looks like the only way to do it to me too.

jedwards Over a year ago

You could also iterate over the letters_guessed global. eg: for letter in letters_guessed: ...

Aswin Murugesh Over a year ago

but letters_guessed may not be a list. if it is, then each time the program checks for all the letters guessed from the beginning, which is time consuming

Piotr Hajduga Over a year ago

@AswinMurugesh what is it then? plural form suggests some kind of container

Piotr Hajduga Over a year ago

well, ok, it may also be some guessed-letter-count... but then again, the whole task is very ambiguous

Daniel Roseman · Accepted Answer · 2013-04-05 20:23:20Z

1

Actually, though I agree with the commenters that this is terribly unPythonic code, you don't actually need to modify the function to make it work. letter is not assigned to within the function, which means that there's no need to declare it as global there: as long as it's defined in the outer scope, it will be accessible within.

answered Apr 5, 2013 at 20:23

Daniel Roseman

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Comments

Ryan Haining · Accepted Answer · 2013-04-05 20:37:52Z

EDIT if you're allowed to freely change the function, then just do

def print_guessed(letter):
    global letters_guessed
    letters_guessed.append(letter)
    return ''.join(c if c in letters_guessed else '-'
                   for c in secret_word)

OTHERWISE: this is quite a tricky problem given a function that doesn't even come close to doing the right thing, but I think I've figured it out!

secret_word = "answer"
letters_guessed = []
letter = ''

def fix_print_guessed(func):
    def wrapper(guess):
        global letters_guessed
        global letter
        letter = guess
        letters_guessed.append(letter)
        return func()
    return wrapper


@fix_print_guessed
def print_guessed():
    global secret_word
    global letters_guessed
    word = []
    for i in range (0, len(secret_word)):
        if letter == secret_word[i]: word.append(letter)
        else: word.append('-')
    word_str = ''.join(word)
    return word_str

found = len(secret_word) * '-'
while '-' in found:
    guess = raw_input('enter a letter: ')
    res = print_guessed(guess)
    found = ''.join(c1 if c1 != '-' else c2 
                    for c1, c2 in zip(found, res))
    print found

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