为什么实例方法在Python 3中被称为类方法? [英] Why can instance methods be called as class methods in Python 3?

问题描述

请考虑以下课程:

class Foo(object):
    def bar(self):
        print(self)

在Python 2 (2.7.13)中，将bar()作为类方法调用会引发异常:

In Python 2 (2.7.13), calling bar() as a class method raises an exception:

>>> Foo.bar('hello')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unbound method bar() must be called with Foo instance as first argument (got str instance instead)

>>> Foo.bar()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unbound method bar() must be called with Foo instance as first argument (got nothing instead)

当调用bar()作为实例方法时，在不带参数调用时将self识别为实例

When bar() is called as an instance method it recognizes self as the instance when called without arguments

>>> Foo().bar('hello')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: bar() takes exactly 1 argument (2 given)

>>> Foo().bar()
<__main__.Foo object at 0x10a8e1a10>

在Python 3 (3.6.0)中，将bar()作为类方法调用时，第一个参数被接受为self:

In Python 3 (3.6.0), when calling bar() as a class method, the first argument is accepted as self:

>>> Foo.bar('hello')
hello

>>> Foo.bar()
 Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: bar() missing 1 required positional argument: 'self'

调用bar()作为实例方法的工作方式与Python 2中一样

Calling bar() as an instance method works as in Python 2

>>> Foo().bar('hello')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: bar() takes 1 positional argument but 2 were given

>>> Foo().bar()
<__main__.Foo object at 0x104ab34a8>

推荐答案

在Python 3上，Foo.bar只是您编写的bar函数.它使用一个参数(恰好名为self)并将其打印出来.您可以在任何函数上调用该函数，它将打印您传递的任何参数.

On Python 3, Foo.bar is just that bar function you wrote. It takes one parameter, which happens to be named self, and prints it. You can call that function on anything, and it will print whatever argument you pass it.

在Python 2上，Foo.bar并不完全是您编写的bar函数.当您访问Foo.bar时，Python会生成一个包装bar函数的未绑定方法对象.未绑定方法对象的工作原理与bar函数相似，主要区别在于验证其第一个参数是Foo的实例.你可以做

On Python 2, Foo.bar isn't quite the bar function you wrote. When you access Foo.bar, Python generates an unbound method object wrapping the bar function. The unbound method object works mostly like the bar function, with the main difference of validating that its first argument is an instance of Foo. You could do

Foo.bar(some_foo_instance)

的作用类似于some_foo_instance.bar()，但是调用Foo的实现，绕过子类中的所有覆盖.但是，您不能执行Foo.bar('hello').

which would work like some_foo_instance.bar(), but calling Foo's implementation, bypassing any overrides in subclasses. You couldn't do Foo.bar('hello'), though.

Python 3删除了未绑定的方法对象.它们不再存在.这使语言更简单，但它删除了用于执行的验证未绑定方法对象.

Python 3 removed unbound method objects. They don't exist any more. That makes the language a bit simpler, but it removes the validation unbound method objects used to perform.

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