python while循环意外行为

Question

I'm relatively new to Python, and I don't understand the following code produces the subsequently unexpected output:

x = input("6 divided by 2 is")
while x != 3:
    print("Incorrect. Please try again.")
    x = input("6 divided by 2 is")
    print(x)

the output of which is:

6 divided by 2 is 3
Incorrect. Please try again.
6 divided by 2 is 3
3
Incorrect. Please try again.
6 divided by 2 is 

Why is the while loop still being executed even though x is equal to 3?

Answer 1

input() returns a string, which you are comparing to an integer. This will always return false. You'll have to wrap input() in a call to int() for a valid comparison.

x = int(input("6 divided by 2 is"))
while x != 3:
    print("Incorrect. Please try again.")
    x = int(input("6 divided by 2 is"))
    print(x)

Read more on int() here .

Answer 2

You are getting this error is because you are not parsing the input like so:

x = int(input("6 divided by 2 is"))

If you replace your inputer statement with that, it'll work.

Answer 3

input method gives the string. So you need to typecast to int as:

x = int(input("6 divided by 2 is"))

Answer 4

Here is my answer to your question

Guesses = 0
while(Guesses < 101):
    try:
        x = int(input("6 divided by 2 is: "))
        if(x == 3):
            print("Correct! 6 divide by 2 is", x)
            break
        else:
            print("Incorrect. try again")
            Guesses += 1
    except ValueError:
        print("That is not a number. Try again.")
        Guesses += 1
else:
    print("Out of guesses.")

I am assuming you wanted the user to input a number so i put your code into a while\\else loop containing a try\\except loop . The try\\except loop ensures that if the users inputs a number, a ValueError will appear and it will inform them that what they inputted was not a number . The while\\else loop ensures that the user will be inputted the question if the Guesses limit is no higher than 100 . This code will ensure that if the user guesses the answer which is 3 , the user will be prompted that they got the answer right and the loop will end ; if the users guesses anything besides 3 (Number wise) the answer will be incorrect and the user will be prompted the question again ; if the user guesses a string it will be classified as a ValueError and they will be notified that their answer wasn't a number and that the user has to try again .

Considering this was asked a long time ago, I'm assuming your probably forgot about this or you figured out the answer but if not try this and tell me if you like this answer. Thank :)

Answer 5

I actually tried this myself now with python 2.6, and did get an int without converting to int. For example, when I try the following:

x = input("6 divided by 2 is")
print "Your input is %s, %s" % (x, type(x))

I get the following:

6 divided by 2 is 2
Your input is 2, <type 'int'>

So is this a version issue? Or maybe an environment issue (I'm using OS X)? What I do conclude is that it should be a good practice using previous recommendations using int().