Compound Assignment To Python Class And Instance Variables
Solution 1:
Personally, I've found these documents by Shalabh Chaturvedi extremely useful and informative regarding this subject matter.
bar.num += 1
is a shorthand for bar.num = bar.num + 1
. This is picking up the class variable Foo.num
on the righthand side and assigning it to an instance variable bar.num
.
Solution 2:
In the following code, num
is a class member.
classFoo:
num = 0
A C++ equivalent would be something like
structFoo {
staticint num;
};
int Foo::num = 1;
classFoo:def__init__(self):
self.num = 0
self.num
is an instance member (self
being an instance of Foo
).
In C++, it would be something like
structFoo {
int num;
};
I believe that Python allows you to have a class member and an instance member sharing the same name (C++ doesn't). So when you do bar = Foo()
, bar
is an instance of Foo
, so with bar.num += 1
, you increment the instance member.
Solution 3:
bar.num += 1
creates a new instance variable 'num' on the 'bar' instance because it doesn't yet exist (and then adds 1 to this value)
an example:
classFoo:
def__init__(self):
self.num= 1
bar = Foo()
print bar.num
this prints 1
print bar.foo
this gives an error as expected: Traceback (most recent call last): File "", line 1, in AttributeError: Foo instance has no attribute 'foo'
bar.foo = 5
print bar.foo
now this prints 5
so what happens in your example: bar.num is resolved as Foo.num because there's only an class attribute. then foo.num is actually created because you assign a value to it.
Solution 4:
Searching for this very question, both this StackOverflow question and two (rather old, but valid) slides by Guido van Rossum (1, 2) came up high. Guido's slides state this behavior has to do with Python's search order for accessing the attribute (in the case of the example above num
). Thought it'd be nice to put the two together right here :)
Solution 5:
I think you just found a bug in Python there. bar.num += 1 should be an error, instead, it is creating an attribute num in the object bar, distinct from Foo.num.
It's a really strange behavior.
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