dynamic_scoping

__author__ = 'Neys'

"""
Create a function add(n) which returns a function that always adds n to any number

addOne = add(1)
addOne(3) # 4

addThree = add(3)
addThree(3) # 6
"""

# Question is about the principles of closure. AKA a technique for implementing lexically scoped name binding.
"""Lexical scoping: Use of local variables. Local means variables with limited scope, that only exist within a specific
function. Helps avoid name collision between 2 identically named variables.

So what does it mean to be within a function? If a variable is within the scope of a certain function then its scope
is effectively the block of code following the function definition. In that block the variable name exists, and is bound
to an assigned value. Outside the block the variable name does not exist.

Dynamic Scoping:
If the variable name's scope is in a certain function, then it's scope is the time frame with which the function is
 executing. (While function is running the variable name exists and is bound to a value.) After the function returns the
 variable name does not exist. """

pi = 3.1415  # create variable pi in global scope
pi_holder = None

def create_area():
    pi_holder = pi  # local variable, different from pi_holder outside

    def area(r):
        return pi_holder*r*r

    return area  # area() gives error

area = create_area()

print("area(10) = ", area(10))
del pi_holder  # delete global pi_holder w/o mutating local lexical pi_holder

try:
    pi_holder # proves pi_holder is created and removed within scope of create_area() and is local to area(r)
except NameError:
    print("pi_holder is empty")

print("area(100) = ", area(100))


x = 42
def local_var_test():
    temp = x  # correct variable assignment
    print("LHS temp assignment to global variable x = ",temp)  # 42
#   x = temp  # incorrect because x is referenced before assignment to local scope of x
#   print(x)  # incorrect also

n = None  # define n

def add(n=n):  # n is what it is arg passed upon function
    def add_to_n(k):
        return k + n
    return add_to_n

addOne = add(1)
addTwo = add(2)
addZero= add(0)
print(addZero(7))

# Final Solution!
def add(n):
    def add_to_x(x):
        return x + n
    return add_to_x