问题
I would like to subclass an immutable type or implement one of my own which behaves like an int
does as shown in the following console session:
>>> i=42
>>> id(i)
10021708
>>> i.__iadd__(1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'int' object has no attribute '__iadd__'
>>> i += 1
>>> i
43
>>> id(i)
10021696
Not surprisingly, int
objects have no __iadd__()
method, yet applying +=
to one doesn't result in an error, instead it apparently creates a new int
and also somehow magically reassigns it to the name given in the augmented assignment statement.
Is it possible to create a user-defined class or subclass of a built-in immutable one that does this, and if so, how?
回答1:
The return value of __iadd__()
is used. You don't need to return the object that's being added to; you can create a new one and return that instead. In fact, if the object is immutable, you have to.
import os.path
class Path(str):
def __iadd__(self, other):
return Path(os.path.join(str(self), str(other)))
path = Path("C:\\")
path += "windows"
print path
回答2:
Simply don't implement __iadd__
, but only __add__
:
>>> class X(object):
... def __add__(self, o):
... return "added"
>>> x = X()
>>> x += 2
>>> x
'added'
If there's no x.__iadd__
, Python simply calculates x += y
as x = x + y
doc.
回答3:
When it sees i += 1
, Python will try to call __iadd__
. If that fails, it'll try to call __add__
.
In both cases, the result of the call will be bound to the name, i.e. it'll attempt i = i.__iadd__(1)
and then i = i.__add__(1)
.
回答4:
class aug_int:
def __init__(self, value):
self.value = value
def __iadd__(self, other):
self.value += other
return self
>>> i = aug_int(34)
>>> i
<__main__.aug_int instance at 0x02368E68>
>>> i.value
34
>>> i += 55
>>> i
<__main__.aug_int instance at 0x02368E68>
>>> i.value
89
>>>
来源:https://stackoverflow.com/questions/11836570/how-to-implement-iadd-for-an-immutable-type