Question

我编写了一个程序来计算一个数字的阶乘，所有工作都完美无缺，但是当测试我输入一个浮点数时崩溃了。我的目标是接受浮点数但不计算。由于程序将接受但返回类似“错误条目，只接受整数”之类的东西。我尝试了多个语句但它只适用于我在语句中输入的数字。所以我想也许应该建立一些东西，可能是通过命名一些浮点数并进行某种减法。但我迷路了。这是我迄今为止没有包含浮动语句的程序：

    def main():
# take input from the user
        num = int(input("Enter a number: "))
        factorial = 1
        if num > 100:
            print("Bad entry. It should be an integer less than or equal to 100!")
            print("Please try again: ")
        elif num == 0:
            print("The factorial of 0 is 1")
        elif num < 0:
            print("Bad entry. It should be an integer superior than or equal to 0!")
            print("Please try again: ")  
        else:
            for i in range(1,num + 1):
                factorial = factorial*i
            print("The factorial of",num,"is",factorial)

main()

Answer 1

您应该使用try / catch块，因为int('3.2')（或任何其他浮点字符串）将引发错误。例如：

try: num = int(input('Enter a number...'))
except ValueError:
   print 'We only accept integers.'
   return

Answer 2

正如许多人所建议的那样，你应该使用try/except块。但是，如果您想接受"6.12"之类的用户输入，并且只是从整数部分计算，您应该这样做：

user_in = "6.12" # or whatever the result from the input(...) call is
user_in = int(float(user_in)) # 6

int不能对不是整数的字符串进行操作，但它可以在浮点数上运行。在字符串上调用float将为您提供一个浮点数，并且在该浮点数上调用int将返回整数部分。

Answer 3

def main():
    # take input from the user
    num = float(input("Enter a number: "))
    if (num%1 != 0):
        print("Bad entry, only integers are accepted.")
        return 

    num = int(num)
    factorial = 1
    if num > 100:
        print("Bad entry. It should be an integer less than or equal to 100!")
        print("Please try again: ")
    elif num == 0:
        print("The factorial of 0 is 1")
    elif num < 0:
        print("Bad entry. It should be an integer superior than or equal to 0!")
        print("Please try again: ")  
    else:
        for i in range(1,num + 1):
            factorial = factorial*i
        print("The factorial of",num,"is",factorial)

main()

如何阻止我的python程序崩溃

3 个答案: