[英]Define operator for built-in and class instance
In Python you can override an operation for your class (say, addition) by defining __add__
. 在Python中,您可以通过定义
__add__
来覆盖类的操作(例如加法)。 This will make it possible add class instance with other values/instances, but you can't add built-ins to instances: 这将使添加具有其他值/实例的类实例成为可能,但是您不能将内置实例添加到实例中:
foo = Foo()
bar = foo + 6 # Works
bar = 6 + foo # TypeError: unsupported operand type(s) for +: 'int' and 'Foo'
Is there any way to make enable this? 有什么办法可以做到这一点?
当实例在右侧时__radd__(self, other)
必须定义方法__radd__(self, other)
来覆盖运算符+
。
You cannot override the + operator for integers. 您不能为整数覆盖+运算符。 What you should do is to override the
__radd__(self, other)
function in the Foo class only . 您应该做的是仅覆盖Foo类中的
__radd__(self, other)
函数。 The self
variable references a Foo
instance, not an integer, and the other
variable references the object on the left side of the + operator. self
变量引用一个Foo
实例,而不是整数, other
变量引用+运算符左侧的对象。 When bar = 6 + foo
is evaluated, the attempt to evaluate 6.__add__(foo)
fails and then Python tries foo.__radd__(6)
(reverse __add__
). 当计算
bar = 6 + foo
,尝试评估6.__add__(foo)
失败,然后Python尝试foo.__radd__(6)
(反向__add__
)。 If you override __radd__
inside Foo
, the reverse __add__
succeeds, and the evaluation of 6 + foo
is the result of foo.__radd__(6)
. 如果您在
Foo
重写__radd__
,则反向__add__
成功,并且对6 + foo
的求值是foo.__radd__(6)
。
def __radd__(self, other):
return self + other
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