Whats wrong with my if else code?

Question

Can someone shed some light on what I'm doing wrong with my code? I'm using Python 3.6. And a beginner. Thanks!

import random

dice1 = random.randint(1, 2)
user_input = input("Guess a number: ")

if dice1 == user_input:
    print("Well Done!")
else:
    print("Try again")

@Li357 in this case when entered a number `user_input` will be int though — user1767754, Dec 19 '17 at 05:42
No it is a string until it is explictly converted (in python 3) — Xero Smith, Dec 19 '17 at 05:44
Possible duplicate of [How can I read inputs as integers?](https://stackoverflow.com/questions/20449427/how-can-i-read-inputs-as-integers) — ndrwnaguib, Dec 19 '17 at 05:44
`Python2` it returns int and `Python3` string. :D We were both right — user1767754, Dec 19 '17 at 05:47

user1767754 · Answer 1 · 2017-12-19T06:48:47.553

What you are really looking for is a while loop that keeps asking for the hidden number as long as you have the wrong answer. As the user Li357 stated, in Python3 the input is always a string, so you have to convert it to an int. In Python2 you wouldn't have to put the int (just in this specific case)

import random

dice1 = random.randint(1, 2)
user_input = None


while(dice1 != user_input):                          #Keep asking
    user_input = int(input("Guess a number: "))      #Input to int
    if int(user_input) == dice1:                     #Now you check int vs int
        print("Well Done!")
        break                                        #If found, break from loop
    else:
        print("Try Again!")

score 0 · Accepted Answer · edited Jun 20 '20 at 09:12

The input() returns a string, so you have a string in user_input. In dice1, you have an integer. Try print(type(user_input)) and print(type(dice1)). You cannot compare values of different types.

Convert the value in user_input to an int, then compare it with dice1, like so:

import random

dice1 = random.randint(1, 2)
user_input = input("Guess a number: ")
user_input = int(user_input)
# You could replace the above two user_input statements with:
# user_input = int(input("Guess a number: ")) # Uncomment before using

if dice1 == user_input:
    print("Well Done!")
else:
    print("Try again")

Running the above code:

Guess a number: 1
Well Done!
>>>

More about input():

input([prompt])

If the prompt argument is present, it is written to standard output without a trailing newline. The function then reads a line from input, converts it to a string (stripping a trailing newline), and returns that.

Thank you very much srig. That corrected my problem. – 000 Dec 20 '17 at 04:40 — 000, Dec 20 '17 at 04:40

score -1 · Answer 3 · answered Dec 19 '17 at 05:41

-1

convert the input to an integer:

import random

dice1 = random.randint(1, 2)
user_input = int(input("Guess a number: "))

if dice1 == user_input:
    print("Well Done!")
else:
    print("Try again")

answered Dec 19 '17 at 05:41

Xero Smith

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please explain the rationale for the down vote. This works as the op intended. The op is using python 3.6 – Xero Smith Dec 19 '17 at 05:52

Whats wrong with my if else code?

3 Answers3