CODEWITHSUNDEEP
python
Justin
Justin

Reputation: 303

Condition always evaluates to else branch

When I type play, a random number is assigned to number1. It asks me for a prediction and I put in a number, say 5. After putting in 5 I always get the else statement and not the if statement. I even put a print() to find out what number was generated. Sometimes I'm right on or within 1 (The game also allows for within 1) and it still re-directs me to the else statement. Could anyone help? Thanks.

money = 1000000

def luckyrollgame():
    global money
    from random import choice
    print('You are in the game lobby of Lucky Roll.')
    print('Choose either \'rules,\' \'play,\' or \'back\'')
    lobby = input()
    if lobby == 'rules':
        luckyrollgamerules()
    if lobby == 'play':
        die = [1, 2, 3, 4, 5, 6]
        number1 = choice(die)
        prediction = input('Please type your prediction number: ')
        if prediction == number1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        if prediction == number1 - 1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        if prediction == number1 + 1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        else:
            print('I\'m sorry. You didn\'t get the number right.')
            print('The number was ' + str(number1) + '.')
            money = money - 1
            print('You now have ' + str(money) + 'dollars.')
            print('--------------------------------------------------')
            altluckyrollgame()
    if lobby == 'back':
        altvillagescene()
    else:
        print('Please type a valid option.')
        print('--------------------------------')
        altluckyrollgame()

*Functions such as altluckyrollgame() or altvillagescene() are part of the game logic and defined elsewhere, so you can ignore them.

Upvotes: 0

Views: 104

Answers (5)

zhangyangyu
zhangyangyu

Reputation: 8610

The prediction returned by input() is a string, so all the comparisons fail. Try casting the value to an integer:

prediction = int(input())

Upvotes: 0

bsoist
bsoist

Reputation: 785

Your have three different if structures there. You almost certainly want

if ...

elif ...

elif ...

else

Upvotes: 0

Travis DePrato
Travis DePrato

Reputation: 402

Use the elif statement after the very first statement. At current, your code

    if lobby == 'back':
        altvillagescene()
    else:
        print('Please type a valid option.')
        print('--------------------------------')
        altluckyrollgame()

is checking if lobby == 'back' and running the else in all other cases. You probably don't want this, as the code under else is run in addition to every other if case.

if x == 0: pass
elif x == 1: pass
else: pass

Code should look like this

money = 1000000

def luckyrollgame():
    global money
    from random import choice
    print('You are in the game lobby of Lucky Roll.')
    print('Choose either \'rules,\' \'play,\' or \'back\'')
    lobby = input()
    if lobby == 'rules':
        luckyrollgamerules()
    elif lobby == 'play':
        die = [1, 2, 3, 4, 5, 6]
        number1 = choice(die)
        prediction = input('Please type your prediction number: ')
######################### This too
        try: prediction = int(prediction)
        except ValueError: prediction = -10
#########################
        if prediction == number1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        elif prediction == number1 - 1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        elif prediction == number1 + 1:
            print('Good job! You guessed right!')
            money = money + 3
            print('You now have ' + str(dollars) + 'dollars.')
        else:
            print('I\'m sorry. You didn\'t get the number right.')
            print('The number was ' + str(number1) + '.')
            money = money - 1
            print('You now have ' + str(money) + 'dollars.')
            print('--------------------------------------------------')
            altluckyrollgame()
    elif lobby == 'back':
        altvillagescene()
    else:
        print('Please type a valid option.')
        print('--------------------------------')
        altluckyrollgame()

Upvotes: 1

Peter Shinners
Peter Shinners

Reputation: 3776

The "else" block is only matched with the final "if prediction == number1 + 1". This means that if the correct number is guessed (or the number1 - 1 case) then it still also run the final else block.

You need to change the code to use "elif" for the intermediate conditionals:

if prediction == number1:
   pass # do the win
elif prediction == number1 - 1
   pass # do the win
elif prediction == number1 + 1
   pass  # do the win
else:
   pass # do the lose

Upvotes: 0

Collin
Collin

Reputation: 12287

Your problem is that you're comparing a string to an integer.

You'll need to first convert the input to an int:

try:
    guess = int(prediction)
except ValueError:
    #Handle when a person enters an invalid number here

Upvotes: 1

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