为什么不能在子类的__init__中传递* args和** kwargs

Question

为了理解* args和** kwargs我做了一些搜索，当我遇到这个问题*args and **kwargs?

所选答案下面的答案引起了我的注意，这就是：

class Foo(object):
    def __init__(self, value1, value2):
    # do something with the values
        print value1, value2

class MyFoo(Foo):
    def __init__(self, *args, **kwargs):
    # do something else, don't care about the args
        print 'myfoo'
        super(MyFoo, self).__init__(*args, **kwargs)

我在这个例子上尝试了一些东西并以这种方式运行代码：

class Foo(object):
    def __init__(self, value1, value2):
    # do something with the values
        print 'I think something is being called here'
        print value1, value2


class MyFoo(Foo):
    def __init__(self, *args, **kwargs):
    # do something else, don't care about the args
        print args, kwargs
        super(MyFoo, self).__init__(*args, **kwargs)


foo = MyFoo('Python', 2.7, stack='overflow')

我明白了：

[...]
    super(MyFoo, self).__init__(*args, **kwargs)
TypeError: __init__() got an unexpected keyword argument 'stack'

更改为super(MyFoo, self).__init__(args, kwargs)

结果是：

('Python', 2.7) {'stack': 'overflow'}
I think something is being called here
('Python', 2.7) {'stack': 'overflow'}

出于某些打击思维的原因，我在质疑这个问题：上面的例子中有什么可能是对的？什么是允许做什么和现实生产中不会做什么？

Answer 1

您的Foo.__init__() 不支持任意关键字参数。您可以将**kw添加到其签名中，以使其接受：

class Foo(object):
    def __init__(self, value1, value2, **kw):
       print 'I think something is being called here'
       print value1, value2, kw

关键字参数仅与具有完全匹配关键字名称的参数匹配;您的Foo方法需要有Python和stack个关键字参数。如果找不到匹配的关键字参数但**kw参数 ，则会在该参数中收集它们。

如果你的子类知道父类只有位置参数，你总是可以传入位置：

class MyFoo(Foo):
    def __init__(self, *args, **kwargs):
    # do something else, don't care about the args
        print args, kwargs
        while len(args) < 2:
            args += kwargs.popitem()
        super(MyFoo, self).__init__(*args[:2])

您现在必须将两个或多个参数传递给MyFoo以使调用工作。

实质上，super().methodname返回对绑定方法的引用;从那里开始它是一个普通方法，所以你需要传入任何方法可以接受的参数。如果您的方法不接受关键字参数，则会出现异常。

Answer 2

原因是所有的参数都已经解压缩为kwargs并且它现在是一个dict。并且您正试图将其传递给正常变量。

def bun(args,kwargs):
print 'i am here'
print kwargs

def fun(*args,**kwargs):
print kwargs
bun(*args,**kwargs)

 fun(hill=3,bi=9) # will fail.


def bun(*args,**kwargs):
print 'i am here'
print kwargs

def fun(*args,**kwargs):
print kwargs
bun(*args,**kwargs) # will work.



fun(hill=3,bi=9)

尝试在

进行修改

class Foo(object):
    def __init__(self, *value1, **value2):
# do something with the values
        print 'I think something is being called here'
        print value1, value2


class MyFoo(Foo):
    def __init__(self, *args, **kwargs):
# do something else, don't care about the args
        print args, kwargs
        super(MyFoo, self).__init__(*args, **kwargs)


foo = MyFoo('Python', 2.7, stack='overflow'

应该工作..！

Answer 3

执行此操作时：

super(MyFoo, self).__init__(*args, **kwargs)

就像你这样做一样，基于代码的工作方式：

super(MyFoo, self).__init__("python", 2.7, stack="overflow")

然而，Foo的 init 函数（MyFoo继承的函数）不支持名为“stack”的关键字参数。

Answer 4

我认为值得补充的是，这可以用来简化子类中的__init__签名。位置参数从左到右被剥离，因此如果将它们添加到前面并将其余部分传递给args和kwargs，则可以避免错误而忘记将它们显式添加到每个子节点。有一些关于这是否是可接受的例外的讨论“显性优于隐式”here。对于深层次结构中的args的长列表，这可能更清晰，更容易维护。

要修改此示例，我将not_for_Foo添加到MyFoo的前面，并将其余内容传递给super。

class Foo(object):
    def __init__(self, a_value1, a_value2, a_stack=None, *args, **kwargs):
        """do something with the values"""
        super(Foo, self).__init__(*args, **kwargs) # to objects constructor fwiw, but object.__init__() takes no args
        self.value1 = a_value1
        self.value2 = a_value2
        self.stack = a_stack
        return

    def __str__(self):
        return ', '.join(['%s: %s' % (k, v) for k, v in self.__dict__.items()])


class MyFoo(Foo):
    def __init__(self, not_for_Foo, *args, **kwargs):
        # do something else, don't care about the args
        super(MyFoo, self).__init__(*args, **kwargs)
        self.not_for_Foo = not_for_Foo # peals off
        self.myvalue1 = 'my_' + self.value1 # already set by super


if __name__ == '__main__':

    print 'Foo with args'
    foo = Foo('Python', 2.7, 'my stack')
    print foo

    print '\nMyFoo with kwargs'
    myfoo = MyFoo('my not for foo', value2=2.7, value1='Python', stack='my other stack')
    print myfoo


$ python argsNkwargs.py 
Foo with args
value2: 2.7, value1: Python, stack: my stack

MyFoo with kwargs
myvalue1: my_Python, not_for_Foo: my not for foo, value2: 2.7, value1: 
Python, stack: my other stack

-lrm