如何獲得__add__調用

Question

class C(object): 
    def __init__(self, value): 
    self.value = value 

    def __add__(self, other): 
    if isinstance(other, C): 
     return self.value + other.value 
    if isinstance(other, Number): 
     return self.value + other 
    raise Exception("error") 


c = C(123) 

print c + c 

print c + 2 

print 2 + c 

顯然，前兩個打印語句將工作，第三個失敗，因爲int。 add（）不能處理一個C類實例。

246 
125 
    print 2 + c 
TypeError: unsupported operand type(s) for +: 'int' and 'C' 

有沒有辦法來解決這個問題，所以2 + C會導致C. 添加（）被調用？

Answer 1

您需要添加__radd__以及處理相反的情況：

def __radd__(self, other): 
    if isinstance(other, C): 
     return other.value + self.value 
    if isinstance(other, Number): 
     return other + self.value 
    return NotImplemented 

，並請注意您應該不提例外;改爲返回NotImplemented單身人士。這樣其他對象仍然可以嘗試支持__add__或__radd__爲您的對象，並將有機會實現添加。

當您嘗試添加兩種類型a和b時，Python首先嚐試呼叫a.__add__(b);如果該呼叫返回NotImplemented，則嘗試使用b.__radd__(a)。

演示：

>>> from numbers import Number 
>>> class C(object): 
...  def __init__(self, value): 
...   self.value = value 
...  def __add__(self, other): 
...   print '__add__ called' 
...   if isinstance(other, C): 
...    return self.value + other.value 
...   if isinstance(other, Number): 
...    return self.value + other 
...   return NotImplemented 
...  def __radd__(self, other): 
...   print '__radd__ called' 
...   if isinstance(other, C): 
...    return other.value + self.value 
...   if isinstance(other, Number): 
...    return other + self.value 
...   return NotImplemented 
... 
>>> c = C(123) 
>>> c + c 
__add__ called 
246 
>>> c + 2 
__add__ called 
125 
>>> 2 .__add__(c) 
NotImplemented 
>>> 2 + c 
__radd__ called 
125 

Answer 2

您需要在課上實施__radd__。

def __radd__(self, other): 
    return self.value + other 

這會導致自動調用，因爲int類將引發NotImplemented錯誤