Question

打印＆＃34;抱歉，我不明白你的意见......＆＃34;每次。

如果我删除输入的异常部分只是整数，它就可以正常工作。

以下是代码：

num= input('Please think of a number between 0 and 100!')

a=int(num)

if type(a)=="int":
    low=0
    high=100
    mid= (low+high)/2

    while round(mid) != a:

        print("Is your secret number", round(mid),"?")

        x=input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly.")

        if x=="h":
            high=round(mid)
        elif x=="l":
            low=round(mid)
        elif x=="c":
            break
        else:
            print("Sorry, I did not understand your input.")

        mid=(low+high)/2

    if round(mid) ==a:
        print("Is your secret number", round(mid),"?")

        y=input("Enter 'h' to indicate the guess is too high. Enter 'l' to indicate the guess is too low. Enter 'c' to indicate I guessed correctly.")

    print("Game over. Your secret number was:", round(mid))  

else:

    print("Sorry, I did not understand your input...")

Answer 1

由于type(a) == 'int'部分而无法正常工作。请改用type(a) == int或isinstance(a, int)。话虽如此，该语句没有任何意义，因为int(num)预先在异常无效的情况下引发异常。

# First make sure if the number is even a valid integer before casting.
if num.isdigit():
    a = int(num)
    ...
else:
    print("Sorry, I did not understand your input")

Answer 2

它没有超过第一个if语句

>>>type(1)=='int'
False

您可以使用isinstance（尽管在您的代码中，此时它可能不是整数）

if isinstance(a, int):

猜猜我的号码

2 个答案: