How do nested functions work in Python?

Question

def maker(n):
    def action(x):
        return x ** n
    return action

f = maker(2)
print(f)
print(f(3))
print(f(4))

g = maker(3)
print(g(3))

print(f(3)) # stil         


        

Answer 1

People answered correctly about the closure, that is: the valid value for "n" inside action is the last value it had whenever "maker" was called.

One ease way to overcome this is to make your freevar (n) a variable inside the "action" function, which receives a copy of "n" in the moment it is run:

The easiest way to do this is to set "n" as a parameter whose default value is "n" at the moment of creation. This value for "n" stays fixed because default parameters for a function are stored in a tuple which is an attribute of the function itself (action.func_defaults in this case):

def maker(n):
    def action(x, k=n):
        return x ** k
    return action

Usage:

f = maker(2) # f is action(x, k=2)
f(3)   # returns 3^2 = 9
f(3,3) # returns 3^3 = 27

Answer 2

One use is to return a function that maintains a parameter.

def outer_closure(a):
    #  parm = a               <- saving a here isn't needed
    def inner_closure():
        #return parm
        return a              # <- a is remembered 
    return inner_closure

# set parm to 5 and return address of inner_closure function
x5 = outer_closure(5)
x5()
>5

x6 = outer_closure(6)
x6()
>6

# x5 inner closure function instance of parm persists 
x5()
>5

Answer 3

Because at the time when you create the function, n was 2, so your function is:

def action(x):
    return x ** 2

When you call f(3), x is set to 3, so your function will return 3 ** 2