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Decorator fundamentals (Closure) , how/why are function local arguments reusable without passing them in again? [duplicate]

I have seen and used nested functions in Python, and they match the definition of a closure. So why are they called "nested functions" instead of "closures"?
Are nested functions not closures because they are not used by the external world?
UPDATE: I was reading about closures and it got me thinking about this concept with respect to Python. I searched and found the article mentioned by someone in a comment below, but I couldn't completely understand the explanation in that article, so that is why I am asking this question.

A closure occurs when a function has access to a local variable from an enclosing scope that has finished its execution.
def make_printer(msg):
def printer():
print(msg)
return printer
printer = make_printer('Foo!')
printer()
When make_printer is called, a new frame is put on the stack with the compiled code for the printer function as a constant and the value of msg as a local. It then creates and returns the function. Because the function printer references the msg variable, it is kept alive after the make_printer function has returned.
So, if your nested functions don't
access variables that are local to enclosing scopes,
do so when they are executed outside of that scope,
then they are not closures.
Here's an example of a nested function which is not a closure.
def make_printer(msg):
def printer(msg=msg):
print(msg)
return printer
printer = make_printer("Foo!")
printer() #Output: Foo!
Here, we are binding the value to the default value of a parameter. This occurs when the function printer is created and so no reference to the value of msg external to printer needs to be maintained after make_printer returns. msg is just a normal local variable of the function printer in this context.

The question has already been answered by aaronasterling
However, someone might be interested in how the variables are stored under the hood.
Before coming to the snippet:
Closures are functions that inherit variables from their enclosing environment. When you pass a function callback as an argument to another function that will do I/O, this callback function will be invoked later, and this function will — almost magically — remember the context in which it was declared, along with all the variables available in that context.
If a function does not use free variables it doesn't form a closure.
If there is another inner level which uses free variables -- all previous levels save the lexical environment ( example at the end )
function attributes func_closure in python < 3.X or __closure__ in python > 3.X save the free variables.
Every function in python has the closure attribute, but if there are no free variables, it is empty.
example: of closure attributes but no content inside as there is no free variable.
>>> def foo():
... def fii():
... pass
... return fii
...
>>> f = foo()
>>> f.func_closure
>>> 'func_closure' in dir(f)
True
>>>
NB: FREE VARIABLE IS MUST TO CREATE A CLOSURE.
I will explain using the same snippet as above:
>>> def make_printer(msg):
... def printer():
... print msg
... return printer
...
>>> printer = make_printer('Foo!')
>>> printer() #Output: Foo!
And all Python functions have a closure attribute so let's examine the enclosing variables associated with a closure function.
Here is the attribute func_closure for the function printer
>>> 'func_closure' in dir(printer)
True
>>> printer.func_closure
(<cell at 0x108154c90: str object at 0x108151de0>,)
>>>
The closure attribute returns a tuple of cell objects which contain details of the variables defined in the enclosing scope.
The first element in the func_closure which could be None or a tuple of cells that contain bindings for the function’s free variables and it is read-only.
>>> dir(printer.func_closure[0])
['__class__', '__cmp__', '__delattr__', '__doc__', '__format__', '__getattribute__',
'__hash__', '__init__', '__new__', '__reduce__', '__reduce_ex__', '__repr__',
'__setattr__', '__sizeof__', '__str__', '__subclasshook__', 'cell_contents']
>>>
Here in the above output you can see cell_contents, let's see what it stores:
>>> printer.func_closure[0].cell_contents
'Foo!'
>>> type(printer.func_closure[0].cell_contents)
<type 'str'>
>>>
So, when we called the function printer(), it accesses the value stored inside the cell_contents. This is how we got the output as 'Foo!'
Again I will explain using the above snippet with some changes:
>>> def make_printer(msg):
... def printer():
... pass
... return printer
...
>>> printer = make_printer('Foo!')
>>> printer.func_closure
>>>
In the above snippet, I didn't print msg inside the printer function, so it doesn't create any free variable. As there is no free variable, there will be no content inside the closure. Thats exactly what we see above.
Now I will explain another different snippet to clear out everything Free Variable with Closure:
>>> def outer(x):
... def intermediate(y):
... free = 'free'
... def inner(z):
... return '%s %s %s %s' % (x, y, free, z)
... return inner
... return intermediate
...
>>> outer('I')('am')('variable')
'I am free variable'
>>>
>>> inter = outer('I')
>>> inter.func_closure
(<cell at 0x10c989130: str object at 0x10c831b98>,)
>>> inter.func_closure[0].cell_contents
'I'
>>> inn = inter('am')
So, we see that a func_closure property is a tuple of closure cells, we can refer them and their contents explicitly -- a cell has property "cell_contents"
>>> inn.func_closure
(<cell at 0x10c9807c0: str object at 0x10c9b0990>,
<cell at 0x10c980f68: str object at 0x10c9eaf30>,
<cell at 0x10c989130: str object at 0x10c831b98>)
>>> for i in inn.func_closure:
... print i.cell_contents
...
free
am
I
>>>
Here when we called inn, it will refer all the save free variables so we get I am free variable
>>> inn('variable')
'I am free variable'
>>>

Python has a weak support for closure. To see what I mean take the following example of a counter using closure with JavaScript:
function initCounter(){
var x = 0;
function counter () {
x += 1;
console.log(x);
};
return counter;
}
count = initCounter();
count(); //Prints 1
count(); //Prints 2
count(); //Prints 3
Closure is quite elegant since it gives functions written like this the ability to have "internal memory". As of Python 2.7 this is not possible. If you try
def initCounter():
x = 0;
def counter ():
x += 1 ##Error, x not defined
print x
return counter
count = initCounter();
count(); ##Error
count();
count();
You'll get an error saying that x is not defined. But how can that be if it has been shown by others that you can print it? This is because of how Python it manages the functions variable scope. While the inner function can read the outer function's variables, it cannot write them.
This is a shame really. But with just read-only closure you can at least implement the function decorator pattern for which Python offers syntactic sugar.
Update
As its been pointed out, there are ways to deal with python's scope limitations and I'll expose some.
1. Use the global keyword (in general not recommended).
2. In Python 3.x, use the nonlocal keyword (suggested by #unutbu and #leewz)
3. Define a simple modifiable class Object
class Object(object):
pass
and create an Object scope within initCounter to store the variables
def initCounter ():
scope = Object()
scope.x = 0
def counter():
scope.x += 1
print scope.x
return counter
Since scope is really just a reference, actions taken with its fields do not really modify scope itself, so no error arises.
4. An alternative way, as #unutbu pointed out, would be to define each variable as an array (x = [0]) and modify it's first element (x[0] += 1). Again no error arises because x itself is not modified.
5. As suggested by #raxacoricofallapatorius, you could make x a property of counter
def initCounter ():
def counter():
counter.x += 1
print counter.x
counter.x = 0
return counter

Python 2 didn't have closures - it had workarounds that resembled closures.
There are plenty of examples in answers already given - copying in variables to the inner function, modifying an object on the inner function, etc.
In Python 3, support is more explicit - and succinct:
def closure():
count = 0
def inner():
nonlocal count
count += 1
print(count)
return inner
Usage:
start = closure()
another = closure() # another instance, with a different stack
start() # prints 1
start() # prints 2
another() # print 1
start() # prints 3
The nonlocal keyword binds the inner function to the outer variable explicitly mentioned, in effect enclosing it. Hence more explicitly a 'closure'.

I had a situation where I needed a separate but persistent name space.
I used classes. I don't otherwise.
Segregated but persistent names are closures.
>>> class f2:
... def __init__(self):
... self.a = 0
... def __call__(self, arg):
... self.a += arg
... return(self.a)
...
>>> f=f2()
>>> f(2)
2
>>> f(2)
4
>>> f(4)
8
>>> f(8)
16
# **OR**
>>> f=f2() # **re-initialize**
>>> f(f(f(f(2)))) # **nested**
16
# handy in list comprehensions to accumulate values
>>> [f(i) for f in [f2()] for i in [2,2,4,8]][-1]
16

def nested1(num1):
print "nested1 has",num1
def nested2(num2):
print "nested2 has",num2,"and it can reach to",num1
return num1+num2 #num1 referenced for reading here
return nested2
Gives:
In [17]: my_func=nested1(8)
nested1 has 8
In [21]: my_func(5)
nested2 has 5 and it can reach to 8
Out[21]: 13
This is an example of what a closure is and how it can be used.

People are confusing about what closure is. Closure is not the inner function. the meaning of closure is act of closing. So inner function is closing over a nonlocal variable which is called free variable.
def counter_in(initial_value=0):
# initial_value is the free variable
def inc(increment=1):
nonlocal initial_value
initial_value += increment
print(initial_value)
return inc
when you call counter_in() this will return inc function which has a free variable initial_value. So we created a CLOSURE. people call inc as closure function and I think this is confusing people, people think "ok inner functions are closures". in reality inc is not a closure, since it is part of the closure, to make life easy, they call it closure function.
myClosingOverFunc=counter_in(2)
this returns inc function which is closing over the free variable initial_value. when you invoke myClosingOverFunc
myClosingOverFunc()
it will print 2.
when python sees that a closure sytem exists, it creates a new obj called CELL. this will store only the name of the free variable which is initial_value in this case. This Cell obj will point to another object which stores the value of the initial_value.
in our example, initial_value in outer function and inner function will point to this cell object, and this cell object will be point to the value of the initial_value.
variable initial_value =====>> CELL ==========>> value of initial_value
So when you call counter_in its scope is gone, but it does not matter. because variable initial_value is directly referencing the CELL Obj. and it indirectly references the value of initial_value. That is why even though scope of outer function is gone, inner function will still have access to the free variable
let's say I want to write a function, which takes in a function as an arg and returns how many times this function is called.
def counter(fn):
# since cnt is a free var, python will create a cell and this cell will point to the value of cnt
# every time cnt changes, cell will be pointing to the new value
cnt = 0
def inner(*args, **kwargs):
# we cannot modidy cnt with out nonlocal
nonlocal cnt
cnt += 1
print(f'{fn.__name__} has been called {cnt} times')
# we are calling fn indirectly via the closue inner
return fn(*args, **kwargs)
return inner
in this example cnt is our free variable and inner + cnt create CLOSURE. when python sees this it will create a CELL Obj and cnt will always directly reference this cell obj and CELL will reference the another obj in the memory which stores the value of cnt. initially cnt=0.
cnt ======>>>> CELL =============> 0
when you invoke the inner function wih passing a parameter counter(myFunc)() this will increase the cnt by 1. so our referencing schema will change as follow:
cnt ======>>>> CELL =============> 1 #first counter(myFunc)()
cnt ======>>>> CELL =============> 2 #second counter(myFunc)()
cnt ======>>>> CELL =============> 3 #third counter(myFunc)()
this is only one instance of closure. You can create multiple instances of closure with passing another function
counter(differentFunc)()
this will create a different CELL obj from the above. We just have created another closure instance.
cnt ======>> difCELL ========> 1 #first counter(differentFunc)()
cnt ======>> difCELL ========> 2 #secon counter(differentFunc)()
cnt ======>> difCELL ========> 3 #third counter(differentFunc)()

I'd like to offer another simple comparison between python and JS example, if this helps make things clearer.
JS:
function make () {
var cl = 1;
function gett () {
console.log(cl);
}
function sett (val) {
cl = val;
}
return [gett, sett]
}
and executing:
a = make(); g = a[0]; s = a[1];
s(2); g(); // 2
s(3); g(); // 3
Python:
def make ():
cl = 1
def gett ():
print(cl);
def sett (val):
cl = val
return gett, sett
and executing:
g, s = make()
g() #1
s(2); g() #1
s(3); g() #1
Reason: As many others said above, in python, if there is an assignment in the inner scope to a variable with the same name, a new reference in the inner scope is created. Not so with JS, unless you explicitly declare one with the var keyword.

For the readers of Structure and Interpretation of Computer Programs (SICP): there are 2 unrelated meanings of closure (CS VS Math), see Wikipedia for the latter/less common one:
Sussman and Abelson also use the term closure in the 1980s with a second, unrelated meaning: the property of an operator that adds data to a data structure to also be able to add nested data structures. This usage of the term comes from the mathematics usage rather than the prior usage in computer science. The authors consider this overlap in terminology to be "unfortunate."
https://en.wikipedia.org/wiki/Closure_(computer_programming)#History_and_etymology
The second (mathematical) meaning is also used in SICP in Python, see for example the discussion of tuples
Our ability to use tuples as the elements of other tuples provides a new means of combination in our programming language. We call the ability for tuples to nest in this way a closure property of the tuple data type. In general, a method for combining data values satisfies the closure property if the result of combination can itself be combined using the same method.
2.3 Sequences | SICP in Python

Here is presented a way to identify if a function is a closure or not via code objects.
As already mentioned in other answers, not every nested function is a closure. Given a composite function (which represent the overall action) its intermediate states can be either be a closure or a nested function.
A closure is a kind function which is "parametrized" by its (non-empty) enclosing scope, the space of free-variables. Notice that a composite function may be made by both types.
The (Python's) internal type code
object represents the compiled function body. Its attribute co_cellvars and co_freevars can be used to "lookaround" the closure/scope of a function.
As mentioned in the doc
co_freevars: tuple of names of free variables (referenced via a function’s closure)
co_cellvars: tuple of names of cell variables (referenced by containing scopes).
Once the function is read, by performing recursive calls a partial function is returned with its own __closure__ (hence cell_contents) and a list of free-variables from its clousre and in its scope.
Let introduce some support functions
# the "lookarounds"
def free_vars_from_closure_of(f):
print(f.__name__, 'free vars from its closure', f.__code__.co_cellvars)
def free_vars_in_scopes_of(f):
print(f.__name__, 'free vars in its scope ', f.__code__.co_freevars)
# read cells values
def cell_content(f):
if f.__closure__ is not None:
if len(f.__closure__) == 1: # otherwise problem with join
c = f.__closure__[0].cell_contents
else:
c = ','.join(str(c.cell_contents) for c in f.__closure__)
else:
c = None
print(f'cells of {f.__name__}: {c}')
Here an example from another answer rewritten in a more systematic way
def f1(x1):
def f2(x2):
a = 'free' # <- better choose different identifier to avoid confusion
def f3(x3):
return '%s %s %s %s' % (x1, x2, a, x3)
return f3
return f2
# partial functions
p1 = f1('I')
p2 = p1('am')
# lookaround
for p in (f1, p1, p2):
free_vars_in_scopes_of(p)
free_vars_from_closure_of(p)
cell_content(p)
Output
f1 free vars in its scope () # <- because it's the most outer function
f1 free vars from its closure ('x1',)
cells of f1: None
f2 free vars in its scope ('x1',)
f2 free vars from its closure ('a', 'x2')
cells of f2: I
f3 free vars in its scope ('a', 'x1', 'x2')
f3 free vars from its closure () # <- because it's the most inner function
cells of f3: free, I, am
The lambda counterpart:
def g1(x1):
return lambda x2, a='free': lambda x3: '%s %s %s %s' % (x1, x2, a, x3)
From the point of view of the free variables/scoping are equivalent. The only minor differences are some values of some attributes of the code object:
co_varnames, co_consts, co_code, co_lnotab, co_stacksize... and natuarlly the __name__ attribute.
A mixed example, closures and not at once:
# example: counter
def h1(): # <- not a closure
c = 0
def h2(c=c): # <- not a closure
def h3(x): # <- closure
def h4(): # <- closure
nonlocal c
c += 1
print(c)
return h4
return h3
return h2
# partial functions
p1 = h1()
p2 = p1()
p3 = p2('X')
p1() # do nothing
p2('X') # do nothing
p2('X') # do nothing
p3() # +=1
p3() # +=1
p3() # +=1
# lookaround
for p in (h1, p1, p2, p3):
free_vars_in_scopes_of(p)
#free_vars_from_closure_of(p)
cell_content(p)
Output
1 X
2 X
3 X
h1 free vars in its scope ()
cells of h1: None
h2 free vars in its scope ()
cells of h2: None
h3 free vars in its scope ('c',)
cells of h3: 3
h4 free vars in its scope ('c', 'x')
cells of h4: 3,X
h1 and h2 are both not closures since they have no cell and no free-variables in their scope.
h3 and h3 are closures and share (in this case) the same cell and free-variable for c. h4 has a further free-variable x with its own cell.
Final considerations:
the __closure__ attribute and __code__.co_freevars can be used to check for values and names (identifiers) of the free-variables
anti-analogies (in a broad sense) between nonlocal and __code__.co_cellvars: nonlocal acts towards the outer function, __code__.co_cellvars instead towards the internal function

When do we need nonlocal in a higher order function?

The following code #1 requires nonlocal previous_score
def announce_max(previous_score = 0):
"""
print the max of all scores
>>> f0 = announce_max()
>>> f1 = f0(5)
5
>>> f2 = f1(2)
5
>>> f3 = f2(3)
5
>>> f4 = f3(7)
7
"""
def say(current_score):
nonlocal previous_score
previous_score = max(previous_score, current_score)
print(previous_score)
return announce_max(previous_score)
return say
The following code #2 does NOT require nonlocal previous_score
def announce_max(previous_score = 0):
"""
print the max of all scores
>>> f0 = announce_max()
>>> f1 = f0(5)
5
>>> f2 = f1(2)
5
>>> f3 = f2(3)
5
>>> f4 = f3(7)
7
"""
def say(current_score):
# nonlocal previous_score
print(max(previous_score, current_score))
return announce_max(max(previous_score, current_score))
return say
Why do we need nonlocal previous_score in code #1 above, do we call it outside the function say?
If yes, why code #2 works?

You need it when you assign to the name, and want to change the closure scoped variable. The act of assignment is what makes a name a local in a function (and it is a local from the beginning of the function, it doesn't switch roles at the moment of assignment); nonlocal and global declarations are what undoes that automatic creation of a local name.
So long as you don't assign to it, reading from that name invokes LEGB lookup, logically checking the local scope first, then enclosing scope(s), then global, then built-in scope. In practice, a name's presence in local or enclosing scopes is statically determined, so at most two scopes are ever checked (when it's not local or enclosing, it checks the global scope first, and falls back to built-in scope if a name being read is not in global scope).

The second version of your function do not work as intended - it will not update the maximum previous score as in the first function.
What is relevant is: outter-scope variables are always readable from nested functions, but they cannot be updated.
If your second function would try to update previous_score without the non local declaration, the function would error with NameError in the line you try to read the value in previous_score to calculate the max. That is: if the name is written to in the nested function, Python assumes it to be local to the function, unless it is explicitly declared as nonlocal or as global. In that case, in the following code, the previous_score local to say would not be assigned when calling max:
# this is broken:
def announce_max(previous_score = 0):
def say(current_score):
previous_score = max(previous_score, current_score)
print(previous_score)
return announce_max(previous_score)
return say

Can anybody explain about the closure of function in Python?

Python
Can anyone help me to understand this code, I am new to Python, how does this function work?
def makeInc(x):
def inc(y):
return y + x
return inc
incOne = makeInc(1)
incFive = makeInc(5)
print(incOne(5)) # returns 6
print(incFive(5)) # returns 10

Higher-order functions
Functions like makeInc that in turn, return another function are called higher order functions. Usually, functions are known to accept data as input and return data as output. With higher order functions, functions instead of data, either return code as output or accept code as input. This code is wrapped into a function. In Python, functions are first class citizens which means functions, just like data, can be passed around. For instance:
myvariable = print
Notice, how I have assigned print to myvariable and how I have dropped the parentheses after print Functions without parentheses are called function objects. This means myvariable now is just another name for print:
print("Hello World!")
myvariable("Hello World!")
Both of the above statements do the exact same thing. What can be assigned to variables can also be returned from functions:
def myfunction():
return print
myfunction()("Hello World!");
Now let's look at your example:
def makeInc(x):
def inc(y):
return y + x
return inc
makeInc is a function that accepts a parameter called x. It then defines another nested inner function called inc which takes in a parameter called y. The thing about nested functions is that they have access to the variables of the enclosing function as well. Here, inc is the inner function but it has access to x which is a variable of the enclosing outer scope.
The last statement return inc returns the inner function to the caller of makeInc. What makeInc essentially is doing, is creating a custom function based on the parameter it receives.
For instance:
x = makeInc(10)
makeInc will first accept 10 and then return a function that takes in an argument y and it increments y by 10.
Here, x is a function that takes in any argument y and then increments it by 10:
x(42) # Returns 52
nonlocal
However, there is a caveat when using nested functions:
def outer():
x = 10
def inner():
x = 20
inner()
print(x) # prints 10
Here, you would assume that the last print statement will print 20. But no! When you assign x = 20 in the inner function, it creates a new local variable called x which is initialized to 20. The outer x remains untouched. To modify the outer x, use the nonlocal keyword:
def outer():
x = 10
def inner():
nonlocal x = 20
inner()
print(x) # prints 20
If you are directly reading x inside inner() instead of assigning to it, you do not need nonlocal.

What is happening here is that makeInc() returns a function handle pointing to specific implementation of inc(). So, calling makeInc(5) "replaces" the x in inc(y) to 5 and returns the callable handle of that function. This handle is saved in incFive. You can now call the function as defined (inc(y)). Since you set x=5 before, the result will be y+5.

nested function change variable in an outside function not working [duplicate]

This question already has answers here:
UnboundLocalError trying to use a variable (supposed to be global) that is (re)assigned (even after first use)
(14 answers)
Closed 6 months ago.
def some_func(a):
def access_a():
print(a)
access_a()
outputs the value of a. However, if I want to change a in the nested function like this:
def some_func(a):
def change_a():
a += 1
print(a)
change_a()
it raises UnboundLocalError exception.
I know a is a nonlocal variable, but why can I access it without declaring nonlocal a?

Python scoping rules 101:
a name bound in a function body is considered local unless explicitely declared global (Python 2.x and 3.x) or nonlocal (Python 3.x only). This holds true whereever the assignment happens in the function's body. Trying to read a local variable before it's bound is of course an error.
if a name is read but not bound in a function's body, it will be looked up in enclosing scopes (outer function(s) if any then global scope). NB: functions arguments are de facto local names so they will never be looked up in enclosing scopes.
Note that a += 1 is mainly a shortcut for a = a + 1, so in your example a is local (bound in the function's body and not explicitely declared global or nonlocal), but you try to read it (the rhs of a = a+1) before it's bound.
In Python 3 you can solve this with a nonlocal statement:
>>> def outer(a):
... def change():
... nonlocal a
... a += 1
... print("before : {}".format(a))
... change()
... print ("after : {}".format(a))
...
>>> outer(42)
before : 42
after : 43
Python 2 doesn't have nonlocal so the canonical hack is to wrap the variable in a mutable container (typically a list but any mutable object would do):
>>> def outer(a):
... _a = [a]
... def change():
... _a[0] += 1
... print("before : {}".format(_a[0]))
... change()
... print ("after : {}".format(_a[0]))
...
>>> outer(42)
before : 42
after : 43
which is quite ugly to say the least.
Now while closures are quite handy, they are mostly the functional counterpart of objects : a way to share state between a set of functions while preserving encapsulation of this state, so if you find you have a need for a nonlocal variable perhaps a proper class might be a cleaner solution (though possibly not for your example that doesn't return the inner function but only uses it internally).

i have two solutions for you:
#first one:
# try with list, compound data types dict/list
def some_func(a):
def change_a():
a[0] += 1
print(a[0])
change_a()
some_func([1])
>>> 2
#second one
#reference pointer
from ctypes import *
def some_func_ctypes(a):
def change_a():
a[0] += 1
print a.contents, a[0]
change_a()
i = c_int(1)
pi = pointer(i)
some_func_ctypes(pi)
>>> c_int(2) 2

When you use the += operator, there is an assignment of a new value to a. This turns a to a local in the eyes of the interpreter.

Python closure with side-effects

I'm wondering if it's possible for a closure in Python to manipulate variables in its namespace. You might call this side-effects because the state is being changed outside the closure itself. I'd like to do something like this
def closureMaker():
x = 0
def closure():
x+=1
print x
return closure
a = closureMaker()
a()
1
a()
2
Obviously what I hope to do is more complicated, but this example illustrates what I'm talking about.

You can't do exactly that in Python 2.x, but you can use a trick to get the same effect: use a mutable object such as a list.
def closureMaker():
x = [0]
def closure():
x[0] += 1
print x[0]
return closure
You can also make x an object with a named attribute, or a dictionary. This can be more readable than a list, especially if you have more than one such variable to modify.
In Python 3.x, you just need to add nonlocal x to your inner function. This causes assignments to x to go to the outer scope.

What limitations have closures in Python compared to language X closures?
nonlocal keyword in Python 2.x
Example:
def closureMaker():
x = 0
def closure():
nonlocal x
x += 1
print(x)
return closure