运算符在python中重载,运算符右侧的对象

Question

我最近在python中了解了运算符重载,我想知道以下是否可行.

考虑下面的hypothetica /设法类.

class My_Num(object):
    def __init__(self, val):
        self.val = val
    def __add__(self, other_num):
        if isinstance(other_num, My_Num):
            return self.val + other_num.val
        else:
            return self.val + other_num

我知道上面写的方式,我可以做这样的事情

n1 = My_Num(1)
n2 = My_Num(2)
n3 = 3
print n1 + n2
print n1 + n3

这些将按预期工作.我也知道它目前写的方式我不能这样做

n1 = My_Num(1)
n2 = 2
print 2 + n1

有没有办法解决？我知道这个例子是设计的,但是我有一个应用程序,如果我在运算符重载时,它将非常有用,我定义运算符的类可以出现在运算符的右侧.这在python中可能吗？

Answer 1

是.例如,有__radd__.另外,有没有对__le__(),__ge__()等等,但是乔尔科内特正确地指出,如果只定义__lt__,a > b调用__lt__的功能b,它提供了一个解决方法.

>>> class My_Num(object):
...     def __init__(self, val):
...         self.val = val
...     def __radd__(self, other_num):
...         if isinstance(other_num, My_Num):
...             return self.val + other_num.val
...         else:
...             return self.val + other_num
... 
>>> n1 = My_Num(1)
>>> n2 = 3
>>> 
>>> print n2 + n1
4
>>> print n1 + n2
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: 'My_Num' and 'int'

请注意,至少在某些情况下,执行以下操作是合理的:

>>> class My_Num(object):
...     def __init__(self, val):
...         self.val = val
...     def __add__(self, other_num):
...         if isinstance(other_num, My_Num):
...             return self.val + other_num.val
...         else:
...             return self.val + other_num
...     __radd__ = __add__