CODEWITHSUNDEEP
pythonclass
int3
int3

Reputation: 13201

Compound assignment to Python class and instance variables

I've been trying to understand Python's handling of class and instance variables. In particular, I found this answer quite helpful. Basically it says that if you declare a class variable, and then you do an assignment to [instance].property, you will be assigning to a different variable altogether -- one in a different namespace from the class variable.

So then I considered -- if I want every instance of my class to have a member with some default value (say zero), should I do it like this:

class Foo:
    num = 0

or like this?

class Foo:
    def __init__(self):
        self.num = 0

Based on what I'd read earlier, I'd think that the second example would be initializing the 'right' variable (the instance instead of the class variable). However, I find that the first method works perfectly well too:

class Foo:
    num = 0

bar = Foo()
bar.num += 1 # good, no error here, meaning that bar has an attribute 'num'
bar.num
>>> 1
Foo.num
>>> 0 # yet the class variable is not modified! so what 'num' did I add to just now?

So.. why does this work? What am I not getting? FWIW, my prior understanding of OOP has come from C++, so explanation by analogy (or pointing where it breaks down) might be useful.

Upvotes: 8

Views: 6556

Answers (5)

akaIDIOT
akaIDIOT

Reputation: 9231

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 :)

Upvotes: 1

Stefan De Boey
Stefan De Boey

Reputation: 2394

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:

class Foo:
  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.

Upvotes: 1

Bertrand Marron
Bertrand Marron

Reputation: 22210

In the following code, num is a class member.

class Foo:
    num = 0

A C++ equivalent would be something like

struct Foo {
  static int num;
};

int Foo::num = 1;

class Foo:
    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

struct Foo {
  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.

Upvotes: 2

MattH
MattH

Reputation: 38245

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.

Upvotes: 5

Thiago Chaves
Thiago Chaves

Reputation: 9463

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.

Upvotes: -3

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