python3 中object.__init__() takes no parameters 报错问题

Question

初学python

代码如下：

class IntTuple(tuple):

def __new__(cls, iterable):
    g = (x for x in iterable if isinstance(x,int) and x > 0)
    return super(IntTuple, cls).__new__(cls, g)


def __init__(self, iterable):
    super(IntTuple, self).__init__(iterable)

    

t = IntTuple([1,-1,'abc',6,['x','y'],3])
print(t)

为什么在不实现__init__方法时无报错，而以上代码会出现错误？
忘解答

Answer 1

import inspect
class IntTuple(tuple):
    def __new__(cls, iterable):
        g = (x for x in iterable if isinstance(x,int) and x > 0)
        return super(IntTuple, cls).__new__(cls, g)
    def __init__(self, iterable):
        print(inspect.getargspec(super(IntTuple, self).__init__))    # 打印父类的__init__ 
        # ArgSpec(args=['self'], varargs='args', keywords='kwargs', defaults=None)
        print(iterable)

t = IntTuple([1,-1,'abc',6,['x','y'],3])
print(inspect.getargspec(t.__init__))    # 打印你自己定义的__init__
# ArgSpec(args=['self', 'iterable'], varargs=None, keywords=None, defaults=None)

你会发现tuple父类的__init__根本没有参数传递，从报的错误可以看出，tuple没有实现__init__魔术方法，而是直接继承的object。从传递的参数也可以看出来，iterable还是那个没处理的，而不是从__new__里传递过来的g