What is the Python equivalent of static variables inside a function?


What is the idiomatic Python equivalent of this C/C++ code?

void foo()
    static int counter = 0;
    printf("counter is %dn", counter);

specifically, how does one implement the static member at the function level, as opposed to the class level? And does placing the function into a class change anything?

Asked By: andrewdotnich



Use a generator function to generate an iterator.

def foo_gen():
    n = 0
    while True:
        yield n

Then use it like

foo = foo_gen().next
for i in range(0,10):
    print foo()

If you want an upper limit:

def foo_gen(limit=100000):
    n = 0
    while n < limit:
       yield n

If the iterator terminates (like the example above), you can also loop over it directly, like

for i in foo_gen(20):
    print i

Of course, in these simple cases it’s better to use xrange 🙂

Here is the documentation on the yield statement.

Answered By: gnud

A bit reversed, but this should work:

def foo():
    foo.counter += 1
    print "Counter is %d" % foo.counter
foo.counter = 0

If you want the counter initialization code at the top instead of the bottom, you can create a decorator:

def static_vars(**kwargs):
    def decorate(func):
        for k in kwargs:
            setattr(func, k, kwargs[k])
        return func
    return decorate

Then use the code like this:

def foo():
    foo.counter += 1
    print "Counter is %d" % foo.counter

It’ll still require you to use the foo. prefix, unfortunately.

(Credit: @ony)

Answered By: Claudiu

Other answers have demonstrated the way you should do this. Here’s a way you shouldn’t:

>>> def foo(counter=[0]):
...   counter[0] += 1
...   print("Counter is %i." % counter[0]);
>>> foo()
Counter is 1.
>>> foo()
Counter is 2.

Default values are initialized only when the function is first evaluated, not each time it is executed, so you can use a list or any other mutable object to store static values.

Answered By: Jeremy
_counter = 0
def foo():
   global _counter
   _counter += 1
   print 'counter is', _counter

Python customarily uses underscores to indicate private variables. The only reason in C to declare the static variable inside the function is to hide it outside the function, which is not really idiomatic Python.

Answered By: Dave

You can add attributes to a function, and use it as a static variable.

def myfunc():
  myfunc.counter += 1
  print myfunc.counter

# attribute must be initialized
myfunc.counter = 0

Alternatively, if you don’t want to setup the variable outside the function, you can use hasattr() to avoid an AttributeError exception:

def myfunc():
  if not hasattr(myfunc, "counter"):
     myfunc.counter = 0  # it doesn't exist yet, so initialize it
  myfunc.counter += 1

Anyway static variables are rather rare, and you should find a better place for this variable, most likely inside a class.

Answered By: vincent

Python doesn’t have static variables but you can fake it by defining a callable class object and then using it as a function. Also see this answer.

class Foo(object):
  # Class variable, shared by all instances of this class
  counter = 0

  def __call__(self):
    Foo.counter += 1
    print Foo.counter

# Create an object instance of class "Foo," called "foo"
foo = Foo()

# Make calls to the "__call__" method, via the object's name itself
foo() #prints 1
foo() #prints 2
foo() #prints 3

Note that __call__ makes an instance of a class (object) callable by its own name. That’s why calling foo() above calls the class’ __call__ method. From the documentation:

Instances of arbitrary classes can be made callable by defining a __call__() method in their class.

Answered By: daniels

The idiomatic way is to use a class, which can have attributes. If you need instances to not be separate, use a singleton.

There are a number of ways you could fake or munge “static” variables into Python (one not mentioned so far is to have a mutable default argument), but this is not the Pythonic, idiomatic way to do it. Just use a class.

Or possibly a generator, if your usage pattern fits.

Answered By: Teddy

I personally prefer the following to decorators. To each their own.

def staticize(name, factory):
    """Makes a pseudo-static variable in calling function.

    If name `name` exists in calling function, return it. 
    Otherwise, saves return value of `factory()` in 
    name `name` of calling function and return it.

    :param name: name to use to store static object 
    in calling function
    :type name: String
    :param factory: used to initialize name `name` 
    in calling function
    :type factory: function
    :rtype: `type(factory())`

    >>> def steveholt(z):
    ...     a = staticize('a', list)
    ...     a.append(z)
    >>> steveholt.a
    Traceback (most recent call last):
    AttributeError: 'function' object has no attribute 'a'
    >>> steveholt(1)
    >>> steveholt.a
    >>> steveholt('a')
    >>> steveholt.a
    [1, 'a']
    >>> steveholt.a = []
    >>> steveholt.a
    >>> steveholt('zzz')
    >>> steveholt.a

    from inspect import stack
    # get scope enclosing calling function
    calling_fn_scope = stack()[2][0]
    # get calling function
    calling_fn_name = stack()[1][3]
    calling_fn = calling_fn_scope.f_locals[calling_fn_name]
    if not hasattr(calling_fn, name):
        setattr(calling_fn, name, factory())
    return getattr(calling_fn, name)
Answered By: spenthil

Here is a fully encapsulated version that doesn’t require an external initialization call:

def fn():
    print (fn.counter)

In Python, functions are objects and we can simply add, or monkey patch, member variables to them via the special attribute __dict__. The built-in vars() returns the special attribute __dict__.

EDIT: Note, unlike the alternative try:except AttributeError answer, with this approach the variable will always be ready for the code logic following initialization. I think the try:except AttributeError alternative to the following will be less DRY and/or have awkward flow:

def Fibonacci(n):
   if n<2: return n
   Fibonacci.memo=vars(Fibonacci).setdefault('memo',{}) # use static variable to hold a results cache
   return Fibonacci.memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2)) # lookup result in cache, if not available then calculate and store it

EDIT2: I only recommend the above approach when the function will be called from multiple locations. If instead the function is only called in one place, it’s better to use nonlocal:

def TheOnlyPlaceStaticFunctionIsCalled():
    def Fibonacci(n):
       nonlocal memo  # required in Python3. Python2 can see memo
       if n<2: return n
       return memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2))
    print (Fibonacci(200))
Answered By: Riaz Rizvi

One could also consider:

def foo():
        foo.counter += 1
    except AttributeError:
        foo.counter = 1


  • much pythonic (“ask for forgiveness not permission”)
  • use exception (thrown only once) instead of if branch (think StopIteration exception)
Answered By: rav

Prompted by this question, may I present another alternative which might be a bit nicer to use and will look the same for both methods and functions:

def funccounter(statics, add=1):
    statics.seed += add
    return statics.seed

print funccounter()       #1
print funccounter(add=2)  #3
print funccounter()       #4

class ACircle(object):
    def counter(statics, self, add=1):
        statics.seed += add
        return statics.seed

c = ACircle()
print c.counter()      #1
print c.counter(add=2) #3
print c.counter()      #4
d = ACircle()
print d.counter()      #5
print d.counter(add=2) #7
print d.counter()      #8    

If you like the usage, here’s the implementation:

class StaticMan(object):
    def __init__(self):
        self.__dict__['_d'] = {}

    def __getattr__(self, name):
        return self.__dict__['_d'][name]
    def __getitem__(self, name):
        return self.__dict__['_d'][name]
    def __setattr__(self, name, val):
        self.__dict__['_d'][name] = val
    def __setitem__(self, name, val):
        self.__dict__['_d'][name] = val

def static_var2(name, val):
    def decorator(original):
        if not hasattr(original, ':staticman'):    
            def wrapped(*args, **kwargs):
                return original(getattr(wrapped, ':staticman'), *args, **kwargs)
            setattr(wrapped, ':staticman', StaticMan())
            f = wrapped
            f = original #already wrapped

        getattr(f, ':staticman')[name] = val
        return f
    return decorator
Answered By: Claudiu

Many people have already suggested testing ‘hasattr’, but there’s a simpler answer:

def func():
    func.counter = getattr(func, 'counter', 0) + 1

No try/except, no testing hasattr, just getattr with a default.

Answered By: Jonathan

A static variable inside a Python method

class Count:
    def foo(self):
            self.foo.__func__.counter += 1
        except AttributeError: 
            self.foo.__func__.counter = 1

        print self.foo.__func__.counter

m = Count()
m.foo()       # 1
m.foo()       # 2
m.foo()       # 3
Answered By: wannik
def staticvariables(**variables):
    def decorate(function):
        for variable in variables:
            setattr(function, variable, variables[variable])
        return function
    return decorate

@staticvariables(counter=0, bar=1)
def foo():

Much like vincent’s code above, this would be used as a function decorator and static variables must be accessed with the function name as a prefix. The advantage of this code (although admittedly anyone might be smart enough to figure it out) is that you can have multiple static variables and initialise them in a more conventional manner.

Using an attribute of a function as static variable has some potential drawbacks:

  • Every time you want to access the variable, you have to write out the full name of the function.
  • Outside code can access the variable easily and mess with the value.

Idiomatic python for the second issue would probably be naming the variable with a leading underscore to signal that it is not meant to be accessed, while keeping it accessible after the fact.

Using closures

An alternative would be a pattern using lexical closures, which are supported with the nonlocal keyword in python 3.

def make_counter():
    i = 0
    def counter():
        nonlocal i
        i = i + 1
        return i
    return counter
counter = make_counter()

Sadly I know no way to encapsulate this solution into a decorator.

Using an internal state parameter

Another option might be an undocumented parameter serving as a mutable value container.

def counter(*, _i=[0]):
    _i[0] += 1
    return _i[0]

This works, because default arguments are evaluated when the function is defined, not when it is called.

Cleaner might be to have a container type instead of the list, e.g.

def counter(*, _i = Mutable(0)):
    _i.value += 1
    return _i.value

but I am not aware of a builtin type, that clearly communicates the purpose.

Answered By: kdb

Another (not recommended!) twist on the callable object like https://stackoverflow.com/a/279598/916373, if you don’t mind using a funky call signature, would be to do

class foo(object):
    counter = 0;
    def __call__():
        foo.counter += 1
        print "counter is %i" % foo.counter

>>> foo()()
counter is 1
>>> foo()()
counter is 2
Answered By: lost

Sure this is an old question but I think I might provide some update.

It seems that the performance argument is obsolete.
The same test suite appears to give similar results for siInt_try and isInt_re2.
Of course results vary, but this is one session on my computer with python 3.4.4 on kernel 4.3.01 with Xeon W3550.
I have run it several times and the results seem to be similar.
I moved the global regex into function static, but the performance difference is negligible.

isInt_try: 0.3690
isInt_str: 0.3981
isInt_re: 0.5870
isInt_re2: 0.3632

With performance issue out of the way, it seems that try/catch would produce the most future- and cornercase- proof code so maybe just wrap it in function

Answered By: Keji Li

A little bit more readable, but more verbose (Zen of Python: explicit is better than implicit):

>>> def func(_static={'counter': 0}):
...     _static['counter'] += 1
...     print _static['counter']
>>> func()
>>> func()

See here for an explanation of how this works.

Answered By: warvariuc

Instead of creating a function having a static local variable, you can always create what is called a “function object” and give it a standard (non-static) member variable.

Since you gave an example written C++, I will first explain what a “function object” is in C++. A “function object” is simply any class with an overloaded operator(). Instances of the class will behave like functions. For example, you can write int x = square(5); even if square is an object (with overloaded operator()) and not technically not a “function.” You can give a function-object any of the features that you could give a class object.

# C++ function object
class Foo_class {
        int counter;     
        Foo_class() {
             counter = 0;
        void operator() () {  
            printf("counter is %dn", counter);
   Foo_class foo;

In Python, we can also overload operator() except that the method is instead named __call__:

Here is a class definition:

class Foo_class:
    def __init__(self): # __init__ is similair to a C++ class constructor
        self.counter = 0
        # self.counter is like a static member
        # variable of a function named "foo"
    def __call__(self): # overload operator()
        self.counter += 1
        print("counter is %d" % self.counter);
foo = Foo_class() # call the constructor

Here is an example of the class being used:

from foo import foo

for i in range(0, 5):
    foo() # function call

The output printed to the console is:

counter is 1
counter is 2
counter is 3
counter is 4
counter is 5

If you want your function to take input arguments, you can add those to __call__ as well:

# FILE: foo.py - - - - - - - - - - - - - - - - - - - - - - - - -

class Foo_class:
    def __init__(self):
        self.counter = 0
    def __call__(self, x, y, z): # overload operator()
        self.counter += 1
        print("counter is %d" % self.counter);
        print("x, y, z, are %d, %d, %d" % (x, y, z));
foo = Foo_class() # call the constructor

# FILE: main.py - - - - - - - - - - - - - - - - - - - - - - - - - - - - 

from foo import foo

for i in range(0, 5):
    foo(7, 8, 9) # function call

# Console Output - - - - - - - - - - - - - - - - - - - - - - - - - - 

counter is 1
x, y, z, are 7, 8, 9
counter is 2
x, y, z, are 7, 8, 9
counter is 3
x, y, z, are 7, 8, 9
counter is 4
x, y, z, are 7, 8, 9
counter is 5
x, y, z, are 7, 8, 9
Answered By: IdleCustard

This answer builds on @claudiu ‘s answer.

I found that my code was getting less clear when I always had
to prepend the function name, whenever I intend to access a static variable.

Namely, in my function code I would prefer to write:


instead of


So, my solution is to :

  1. add a statics attribute to the function
  2. in the function scope, add a local variable statics as an alias to my_function.statics
from bunch import *

def static_vars(**kwargs):
    def decorate(func):
        statics = Bunch(**kwargs)
        setattr(func, "statics", statics)
        return func
    return decorate

@static_vars(name = "Martin")
def my_function():
    statics = my_function.statics
    print("Hello, {0}".format(statics.name))


My method uses a class named Bunch, which is a dictionary that supports
attribute-style access, a la JavaScript (see the original article about it, around 2000)

It can be installed via pip install bunch

It can also be hand-written like so:

class Bunch(dict):
    def __init__(self, **kw):
        self.__dict__ = self
Answered By: Pascal T.

Soulution n +=1

def foo():
  foo.__dict__.setdefault('count', 0)
  foo.count += 1
  return foo.count
Answered By: Feca

Other solutions attach a counter attribute to the function, usually with convoluted logic to handle the initialization. This is inappropriate for new code.

In Python 3, the right way is to use a nonlocal statement:

counter = 0
def foo():
    nonlocal counter
    counter += 1
    print(f'counter is {counter}')

See PEP 3104 for the specification of the nonlocal statement.

If the counter is intended to be private to the module, it should be named _counter instead.

Answered By: cbarrick

After trying several approaches I ended up using an improved version of @warvariuc’s answer:

import types

def func(_static=types.SimpleNamespace(counter=0)):
    _static.counter += 1
Answered By: VPfB

Building on Daniel’s answer (additions):

class Foo(object): 
    counter = 0  

def __call__(self, inc_value=0):
    Foo.counter += inc_value
    return Foo.counter

foo = Foo()

def use_foo(x,y):
    if(foo() == 10):


The reason why I wanted to add this part is , static variables are used not only for incrementing by some value, but also check if the static var is equal to some value, as a real life example.

The static variable is still protected and used only within the scope of the function use_foo()

In this example, call to foo() functions exactly as(with respect to the corresponding c++ equivalent) :

stat_c +=9; // in c++
foo(9)  #python equiv

if(stat_c==10){ //do something}  // c++

if(foo() == 10):      # python equiv
  #add code here      # python equiv       

Output :

if class Foo is defined restrictively as a singleton class, that would be ideal. This would make it more pythonic.

Answered By: yash

A global declaration provides this functionality. In the example below (python 3.5 or greater to use the “f”), the counter variable is defined outside of the function. Defining it as global in the function signifies that the “global” version outside of the function should be made available to the function. So each time the function runs, it modifies the value outside the function, preserving it beyond the function.

counter = 0

def foo():
    global counter
    counter += 1
    print("counter is {}".format(counter))

foo() #output: "counter is 1"
foo() #output: "counter is 2"
foo() #output: "counter is 3"
Answered By: Richard Merren

I write a simple function to use static variables:

def Static():
    ### get the func object by which Static() is called.
    from inspect import currentframe, getframeinfo
    caller = currentframe().f_back
    func_name = getframeinfo(caller)[2]
    # print(func_name)
    caller = caller.f_back
    func = caller.f_locals.get(
        func_name, caller.f_globals.get(
    class StaticVars:
        def has(self, varName):
            return hasattr(self, varName)
        def declare(self, varName, value):
            if not self.has(varName):
                setattr(self, varName, value)

    if hasattr(func, "staticVars"):
        return func.staticVars
        # add an attribute to func
        func.staticVars = StaticVars()
        return func.staticVars

How to use:

def myfunc(arg):
    if Static().has('test1'):
        Static().test += 1
        Static().test = 1

    # declare() only takes effect in the first time for each static variable.
    Static().declare('test2', 1)
    Static().test2 += 1
Answered By: 0x262f

Using a decorator and a closure

The following decorator can be used create static function variables. It replaces the declared function with the return from itself. This implies that the decorated function must return a function.

def static_inner_self(func):
    return func()

Then use the decorator on a function that returns another function with a captured variable:

def foo():
    counter = 0
    def foo():
        nonlocal counter
        counter += 1
        print(f"counter is {counter}")
    return foo

nonlocal is required, otherwise Python thinks that the counter variable is a local variable instead of a captured variable. Python behaves like that because of the variable assignment counter += 1. Any assignment in a function makes Python think that the variable is local.

If you are not assigning to the variable in the inner function, then you can ignore the nonlocal statement, for example, in this function I use to indent lines of a string, in which Python can infer that the variable is nonlocal:

def indent_lines():
    import re
    re_start_line = re.compile(r'^', flags=re.MULTILINE)
    def indent_lines(text, indent=2):
        return re_start_line.sub(" "*indent, text)
    return indent_lines

P.S. There is a deleted answer that proposed the same. I don’t know why the author deleted it.

Answered By: Miguel Angelo
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