Python OOP – Is @staticmethod Decorator Needed for Static Methods?

Maybe a bit of example code will help: Notice the difference in the call signatures of foo, class_foo and static_foo:

class A(object):
    def foo(self, x):
        print(f"executing foo({self}, {x})")

    @classmethod
    def class_foo(cls, x):
        print(f"executing class_foo({cls}, {x})")

    @staticmethod
    def static_foo(x):
        print(f"executing static_foo({x})")

a = A()

Below is the usual way an object instance calls a method. The object instance, a, is implicitly passed as the first argument.

a.foo(1)
# executing foo(<__main__.A object at 0xb7dbef0c>, 1)

With classmethods, the class of the object instance is implicitly passed as the first argument instead of self.

a.class_foo(1)
# executing class_foo(<class '__main__.A'>, 1)

You can also call class_foo using the class. In fact, if you define something to be a classmethod, it is probably because you intend to call it from the class rather than from a class instance. A.foo(1) would have raised a TypeError, but A.class_foo(1) works just fine:

A.class_foo(1)
# executing class_foo(<class '__main__.A'>, 1)

One use people have found for class methods is to create inheritable alternative constructors.

With staticmethods, neither self (the object instance) nor cls (the class) is implicitly passed as the first argument. They behave like plain functions except that you can call them from an instance or the class:

a.static_foo(1)
# executing static_foo(1)

A.static_foo('hi')
# executing static_foo(hi)

Staticmethods are used to group functions which have some logical connection with a class to the class.

foo is just a function, but when you call a.foo you don't just get the function, you get a "partially applied" version of the function with the object instance a bound as the first argument to the function. foo expects 2 arguments, while a.foo only expects 1 argument.

a is bound to foo. That is what is meant by the term "bound" below:

print(a.foo)
# <bound method A.foo of <__main__.A object at 0xb7d52f0c>>

With a.class_foo, a is not bound to class_foo, rather the class A is bound to class_foo.

print(a.class_foo)
# <bound method type.class_foo of <class '__main__.A'>>

Here, with a staticmethod, even though it is a method, a.static_foo just returns a good 'ole function with no arguments bound. static_foo expects 1 argument, and a.static_foo expects 1 argument too.

print(a.static_foo)
# <function static_foo at 0xb7d479cc>

And of course the same thing happens when you call static_foo with the class A instead.

print(A.static_foo)
# <function static_foo at 0xb7d479cc>

Whether to use @staticmethod or not depends on what you want to achieve. Ignoring the decorator because there is more to type is a rather silly reason (no offense!) and indicates that you have not understood the concept of a static method in Python!

Static methods are independent of the class and any class instance. They only use the class scope as a namespace. If you omit the @staticmethod decorator, you are creating an instance method that cannot be used without constructing an instance.

Here is a very simple class Foo:

>>> class Foo(object):
...    @staticmethod
...    def foo():
...       print 'foo'
...
...    def bar(self):
...       print 'bar'

Now, Foo.foo() is a static method that can be called directly:

>>> Foo.foo()
foo

Foo.bar() on the other hand is an instance method, that can only be called from instances (objects) of Foo:

>>> Foo.bar()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unbound method foo() must be called with Foo instance as first argument (got nothing instead)
>>> foo = Foo()
>>> foo.bar()
bar

To answer your question: If you want to define a static method, use @staticmethod. Otherwise, don't.

If you have a method that does not use self, and therefore could be written as a static method, ask yourself: Will you ever want to access this function from outside without having an instance? Most of the times, the answer will be: No.