python - ValueError: invalid literal for int() with base 10: 'stop'

4
votes

Every time I try me code it works but when I type in 'stop' it gives me an error:

ValueError: invalid literal for int() with base 10: 'stop'

def guessingGame():
    global randomNum
    guessTry = 3

    while True:
        guess = input('Guess a Number between 1 - 10, You have 3 Tries, or Enter Stop:  ')
        if int(guess) == randomNum:
            print('Correct')
            break

        if int(guess) < randomNum:
            print('Too Low')
            guessTry = guessTry - 1
            print('You have, ' + str(guessTry) + ' Guesses Left')

        if int(guess) > randomNum:
            print('Too High')
            guessTry = guessTry - 1
            print('You have, ' + str(guessTry) + ' Guesses Left')

        if guessTry == 0:
            print('You have no more tries')
            return

        if str(guess) == 'stop' or str(guess) == 'Stop':
            break
3
pythonstringint

3 Answers

12
votes

The string passed to int() should only contain digits:

>>> int("stop")
Traceback (most recent call last):
  File "<ipython-input-114-e5503af2dc1c>", line 1, in <module>
    int("stop")
ValueError: invalid literal for int() with base 10: 'stop'

A quick fix will be to use exception handling here:

def guessingGame():
    global randomNum
    global userScore
    guessTry = 3

    while True:
        guess = input('Guess a Number between 1 - 10, You have 3 Tries, or Enter Stop:  ')
        try:
            if int(guess) == randomNum:
                print('Correct')
                break

            if int(guess) < randomNum:
               print('Too Low')
               guessTry = guessTry - 1
               print('You have, ' + str(guessTry) + ' Guesses Left')

            if int(guess) > randomNum:
                print('Too High')
                guessTry = guessTry - 1
                print('You have, ' + str(guessTry) + ' Guesses Left')

            if guessTry == 0:
                print('You have no more tries')
                return
        except ValueError:
            #no need of str() here
            if guess.lower() == 'stop':
                break
guessingGame()

And you can use guess.lower() == 'stop' to match any uppercase-lowercase combination of "stop":

>>> "Stop".lower() == "stop"
True
>>> "SToP".lower() == "stop"
True
>>> "sTOp".lower() == "stop"
True
1
votes

Here's a more pythonic (Python 3) version. 

def guessing_game(random_num):
    tries = 3
    print("Guess a number between 1 - 10, you have 3 tries, or type 'stop' to quit")

    while True:
        guess = input("Your number: ")
        try:
            guess = int(guess)
        except (TypeError, ValueError):
            if guess.lower()  == 'stop' :
                return
            else:
                print("Invalid value '%s'" % guess)
                continue

        if guess == random_num:
            print('Correct')
            return
        elif guess < random_num:
            print('Too low')
        else:
            print('Too high')

        tries -= 1
        if tries == 0:
            print('You have no more tries')
            return

        print('You have %s guesses left' % tries)
0
votes

You are trying to convert the string "stop" to an integer. That string has no valid representation as an integer, so you get that error. You should put 

if str(guess) == 'stop' or str(guess) == 'Stop':
break

as the first check

Another suggestion is to use lowercase on the input and then check for the lowercase 'stop'. That way you will have to check just once and it will capture either 'Stop', 'STOP', 'sTOp', etc..

if str(guess).lower() == 'stop':
break