我们真的需要在python中使用@staticmethod装饰器来声明静态方法吗? [英] Do we really need @staticmethod decorator in python to declare static method

问题描述

我很好奇为什么我们需要 @staticmethod 装饰器将方法声明为 static.我正在阅读 Python 中的静态方法，我开始知道静态方法可以在不实例化其类的情况下调用.所以我尝试了下面的两个例子，但都一样:

I am curious about why we need the @staticmethod decorator to declare method as static. I was reading about static methods in Python, and I came to know that static method can be callable without instantiating its class. So I tried the two examples below, but both do the same:

class StatMethod:
  def stat():
    print("without Decorator")

class StatMethod_with_decorator:
  @staticmethod
  def stat():
    print("With Decorator")

如果我直接在类上调用 stat() 方法，都会打印/显示以下值:

If I call the stat() method on the class directly, both print/show the values below:

>> StatMethod.stat()
without Decorator
>> StatMethod_with_decorator.stat()
With Decorator

推荐答案

如果您打算尝试从类的实例而不是直接从类调用 @staticmethod ，则需要装饰器

You need the decorator if you intend to try to call the @staticmethod from the instance of the class instead of of the class directly

class Foo():
    def bar(x):
        return x + 5

>>> f = Foo()
>>> f.bar(4)
Traceback (most recent call last):
  File "<pyshell#7>", line 1, in <module>
    f.bar(4)
TypeError: bar() takes 1 positional argument but 2 were given

现在，如果我声明 @staticmethod，self 参数不会作为第一个参数隐式传递

Now if I declare @staticmethod the self argument isn't passed implicitly as the first argument

class Foo():
    @staticmethod
    def bar(x):
        return x + 5

>>> f = Foo()
>>> f.bar(4)
9

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