Question

这是我第一次尝试创建和使用类。当我要求用户输入时发生错误。我收到以下错误：

n1 = Arithmetic.float_input("Enter your First number: ")
TypeError: float_input() missing 1 required positional argument: 'msg'

这是我的代码。

# Define class
class Arithmetic:
    def float_input(self, msg): # Use this function for exception handling during user input
        while True:
            try:
                return float(input(msg))
            except ValueError:
                print("You must enter a number!")
            else:
            break
    def add(self, n1, n2):
        sum1 = n1 + n2
        print(n1,"+" ,n2,"=", sum1)
    def sub(self, n1, n2):
        diff = n1 - n2
        print(n1,"-",n2,"-", diff)
    def mult(self, n1, n2):
        product = n1 * n2
        print(n1,"*",n2, "=", product)
    def div(self, n1, n2):
        if n2 == 0:
            print(n1, "/",n2,"= You cannot divide by Zero")
        else:
            quotient = n1 / n2
            print(n1, "/",n2,"=", quotient)
    def allInOne(self, n1, n2):
        #Store values in dictionary (not required, just excercising dictionary skill)
        res = {"add": add(n1, n2), "sub": sub(n1, n2), "mult": mult(n1, n2), "div": div(n1, n2)}

# Declare variables. Ask user for input and use the exception handling function       
n1 = Arithmetic.float_input("Enter your First number: ")
n2 = Arithmetic.float_input("Enter your Second number: ")

我想念什么？

Answer 1

如果您来自Java背景，那么值得一提的是，除非您需要self提供的状态，否则通常不需要在Python中的类中包装方法。

无论如何，您看到的错误是因为您的方法未标记为@classmethod或@staticmethod，因此需要类的实例，而您只是通过类本身来调用它们（因此不会将隐式实例或类对象作为第一个参数传入。

因此，您的选择是：

1 –创建Arithmetic()的实例并使用它：

arith = Arithmetic()
n1 = arith.float_input("Enter your First number: ")
n2 = arith.float_input("Enter your Second number: ")

2 –将方法标记为静态，例如

@staticmethod
def float_input(prompt):  # note: no `self`

3 –标记方法类方法，例如

@classmethod
def float_input(cls, prompt):  # `cls` is `Arithmetic` (or its subclass) itself

4 –使方法成为没有类的常规函数。

Answer 2

问题是您在调用方法之前没有创建Arithmetic的实例。因为您没有创建对象，所以不会将任何实例传递给self参数。这将导致消息"Enter your first number:"传递给 self 参数，而msg参数为空。

要解决此问题，只需在类名称后使用括号创建一个对象，例如：

# Declare variables. Ask user for input and use the exception handling function       
n1 = Arithmetic().float_input("Enter your First number: ")
n2 = Arithmetic().float_input("Enter your Second number: ")

如果您不是故意创建对象，则可以使用@classmethod类修饰符将类名传递给 self 参数。

# Define class

class Arithmetic:

    @classmethod
    def float_input(class_, msg): # Use this function for exception handling during user input

        while True:
            try:
                return float(input(msg))
            except ValueError:
                print("You must enter a number!")
            else:
                break
    # Code...

n1 = Arithmetic.float_input("Enter your First number: ")
n2 = Arithmetic.float_input("Enter your Second number: ")

还有另一个名为@staticmethod的装饰器。如果使用此装饰器，则可以在不具有算术实例且无需在方法签名中定义 self 的情况下调用该方法。示例：

class Arithmetic:

    @staticmethod
    def float_input(msg): # Use this function for exception handling during user input
        while True:
            try:
                return float(input(msg))
            except ValueError:
                print("You must enter a number!")
            else:
                break
    # Code...

n1 = Arithmetic.float_input("Enter your First number: ")
n2 = Arithmetic.float_input("Enter your Second number: ")

Answer 3

修复为：

# Declare variables. Ask user for input and use the exception handling function       
arithmatic = Arithmetic()
n1 = arithmatic.float_input("Enter your First number: ")
n2 = arithmatic.float_input("Enter your Second number: ")

Answer 4

最好先实例化该类，然后使用它的正确方法，如下所示

n1 = new Arithmetic()

n1.float_input('Enter your First number: ')

如何使用Python类并要求用户输入？

4 个答案: