__add__ returns 超类的实例不是子类

__add__ returns instance of superclass not subclass

当我子class一些class时,说int,并自定义它的__add__方法并调用super().__add__(other)它returns一个int 的实例,而不是我的子 class。我可以通过在 returns 和 int 的每个 super() 调用之前添加 type(self) 来解决这个问题,但这似乎过分了。必须有更好的方法来做到这一点。 floatsfractions.Fractions.

也会发生同样的事情
class A(int):
    def __add__(self, other):
        return super().__add__(other)

x = A()
print(type(x + 1))

输出:
<class 'int'>

预期输出:
<class '__main__.A'>

一种方法是创建一个装饰器,它可以用强制转换包装所需的数学运算:

def wrap_math(c):
    def wrapped(orig):
        return lambda s, o: c(orig(s,o))

    maths = ["__add__", "__sub__"]
    for op in maths:
        func = wrapped(getattr(c, op))
        setattr(c, op, func)

return c

@wrap_math
class Special(int)
    pass

 x = Special(10)
 type(x + 10)

完成您要包装的函数列表,您应该可以开始了。

super() 函数从父 class 调用方法,在本例中是 int。相反,您应该在 __add__ 方法中初始化 class:

class A(int):
    def __add__(self, number):
        return A(self.numerator + number)

x = A(4)
print(type(x + 1))

这可以使用 descriptor 来完成。以下 class 使用特殊方法,当 class 在 class 体内实例化时具有特殊效果。

class SuperCaller:

    def __set_name__(self, owner, name):
        """Called when the class is defined. owner is the class that's being
        defined. name is the name of the method that's being defined.
        """
        method = getattr(super(owner, owner), name)
        def call(self, other):
            # Note that this self shadows the __set_name__ self. They are two
            # different things.
            return type(self)(method(self, other))
        self._call = call

    def __get__(self, instance, owner):
        """instance is an instance of owner."""
        return lambda other: self._call(instance, other)


class A(int):
    __add__ = SuperCaller()


x = A()
print(type(x + 1))

输出:<class '__main__.A'>