Question

我想制作一个简单的平方根计算器。

Class ...

    '' Shared .ctor, called once the first time the class is accessed.
    Shared Sub New()
        dictOfAlgorithm = New Generic.Dictionary(Of String, algorithm)
    End Sub

    Public Shared dictOfAlgorithm As Generic.Dictionary(Of String, algorithm)

End Class

似乎我没有做错任何事情，但是，当我尝试运行程序时，在输入数字后，我遇到以下错误消息：

num = input('Enter a number and hit enter: ')

if len(num) > 0 and num.isdigit():
    new = (num**0.5)
    print(new)
else:
    print('You did not enter a valid number.')

Answer 1

您可以使用此解决方案。这里try和catch能够处理各种输入。所以你的程序永远不会失败。并且因为输入被转换为float。您不会遇到任何类型相关的错误。

try:
    num = float(input('Enter a positive number and hit enter: '))
    if num >= 0:
        new = (num**0.5)
    print(new)

except:
    print('You did not enter a valid number.')

Answer 2

输入函数返回字符串值。所以你需要正确地解析它

num = raw_input('Enter a number and hit enter: ')

if num.isdigit():
  if int(num) > 0:
      new = (int(num)**0.5)
      print(new)
else:
    print('You did not enter a valid number.')

Answer 3

使用Math模块进行简单计算。参考：Math module Documentation.

import math
num = raw_input('Enter a number and hit enter: ')

if num.isdigit():
    num = float(num)
    new = math.sqrt(num)
    print(new)
else:
    print('You did not enter a valid number.')

Python平方根计算器错误

3 个答案: