当我从整数中减去浮点数时(例如 1-2.0
),Python 会进行隐式类型转换(我认为).但是当我使用魔术方法 __sub__
调用我认为是相同的操作时,它突然不再存在了.
When I subtract a float from an integer (e.g. 1-2.0
), Python does implicit type conversion (I think). But when I call what I thought was the same operation using the magic method __sub__
, it suddenly does not anymore.
我在这里缺少什么?当我为自己的类重载运算符时,除了将输入显式转换为我需要的任何类型之外,还有其他方法吗?
What am I missing here? When I overload operators for my own classes, is there a way around this other than explicitly casting input to whatever type I need?
a=1
a.__sub__(2.)
# returns NotImplemented
a.__rsub__(2.)
# returns NotImplemented
# yet, of course:
a-2.
# returns -1.0
a - b
不仅仅是 a.__sub__(b)
.如果 a
无法处理该操作,它也会尝试 b.__rsub__(a)
,在 1 - 2.
的情况下,它是float 的 __rsub__
处理操作.
a - b
isn't just a.__sub__(b)
. It also tries b.__rsub__(a)
if a
can't handle the operation, and in the 1 - 2.
case, it's the float's __rsub__
that handles the operation.
>>> (2.).__rsub__(1)
-1.0
您运行了 a.__rsub__(2.)
,但这是错误的 __rsub__
.您需要右侧操作数的 __rsub__
,而不是左侧操作数.
You ran a.__rsub__(2.)
, but that's the wrong __rsub__
. You need the right-side operand's __rsub__
, not the left-side operand.
减法运算符没有内置隐式类型转换.float.__rsub__
必须手动处理整数.如果您想在自己的运算符实现中进行类型转换,您也必须手动处理.
There is no implicit type conversion built into the subtraction operator. float.__rsub__
has to handle ints manually. If you want type conversion in your own operator implementations, you'll have to handle that manually too.
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