Python functions and variables trouble

Question

I have created this code so far...

def print_slow(str):
    for letter in str:
        sys.stdout.write(letter)
        sys.stdout.flush()
        time.sleep(0.005)

def menu():
    print_slow("-------------[MENU]-------------")
    print(" ")
    print_slow("1) Enter a sentence.")
    print(" ")
    print_slow("2) Find the position of  a word.")
    print(" ")
    print_slow("--------------------------------")
    print(" ")

    print_slow(">>> ")
    choice = str(input(" "))
    print(" ")
    time.sleep(0.5)

    if choice == "1":
        option1()
    if choice == "2":
        option2()

def option1():
    print_slow("Enter sentence")
    sentence = str(input(": "))
    print(" ")
    menu()

def option2():
    if not sentence:
        print_slow("Please enter a sentence first!")
        time.sleep(0.5)
        print(" ")

    else:
        sentenceUppercase = sentence.upper()
        [code goes on...]

Basically when I test it, I press option 2 first and it should give the output 'Please enter a sentence first!', which it does.

I then press option 1 in the menu and it should prompt me to input a sentence (I put 'my name is bob' as a test) and it does.

I then pressed option 2 after inputting the sentence and it should continue with my code - instead it gives the error message 'Please enter a sentence first!'

How can I fix this??

Have you defined sentence as a global variable? Because the code you're showing us should give an error because in the function option2 you're checking if sentence evaluates to True (or False), but, from your code, that sentence seems not to be defined anywhere. In other words, sentence under option1 is a local variable and not the same variable as in option2. — nbro
– nbro, Commented Mar 8, 2017 at 16:11
I've done sentence = " " right at the start of my code at the first line but the error still comes up @nbro — Nil
– Nil, Commented Mar 8, 2017 at 16:13

rainer · Accepted Answer · 2017-03-08 16:15:57Z

2

You're setting a local variable sentence inside function option1. This variable is not visible in option2 since it lives inside option1 only and will be cleaned up once option1 is finished.

If you want to share the variable, you need to define it as global at least in option1:

def option1():
    print_slow("Enter sentence")
    global sentence
    sentence = str(input(": "))
    print(" ")
    menu()

Note, however, that using global variables is usually a sign of bad code quality. In your case, it would make more sense to have option1 return sentence to main, and pass it from main to option2.

edited Mar 8, 2017 at 16:15

answered Mar 8, 2017 at 16:10

rainer

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Comments

Will · Accepted Answer · 2017-03-08 16:12:31Z

1

Your issue is in assigning a value to sentence. Since you are assigning it in a function, when you leave that function's scope, you lose the value. Try using global:

sentence = ''

def option1():
    global sentence              # <-- this maintains its value in global scope
    print_slow("Enter sentence")
    sentence = str(input(": "))
    print(" ")
    menu()

def option2():
    global sentence              # <-- and here
    if not sentence:
        print_slow("Please enter a sentence first!")
        time.sleep(0.5)
        print(" ")
    else:
        sentenceUppercase = sentence.upper()

Or you could pass it back and forth with parameters.

answered Mar 8, 2017 at 16:12

Will

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