Understanding excution order in nested functions Python 3.8

Question

Total newbie herem trying to understand the flow of execution,, look at the following code:

from datetime import datetime
from time import *

def time_decorator(function,a=0):
    a +=1
    print(a)
    def wrapper(*args,**kwargs):
        t1= time()
        result = function(*args,**kwargs)
        t2 = time() - t1
        print("elapsed:", t2)
        return result
    a +=2
    print(a)
    return wrapper
@time_decorator
def add(x, y=10):
    sleep(2)
    return x + y

print (add(10))

The output is like this:

1
3
elapsed: 2.0001702308654785
20
1
3

my question is why the last vaues of 'a' are the same in the last 2 lines??, i was expectin to be increased by two more?...im not understanding the flow of excution of the interpreter, help!!

Um, are you sure that you gave the correct output? Because for me the output is different: Without the `print` function, it's `1` and `3`, and with the print function added, it's `elapsed: ...` and `20`, as expected. Also, decorators are tough; sure it's a good starting topic? — sammy, Dec 01 '19 at 17:57
There is no second `1` and `3`; `a` is only used and printed when the decorator runs, when you define `add`. The flow is as follows: `add` is defined, then `time_decorator` is called with `add` as an argument and it returns a new function that calls `add`; this new function is bound to the name `add`. — chepner, Dec 01 '19 at 18:02

score 0 · Answer 1 · answered Dec 01 '19 at 18:04

Decorators are just a fancy form of function application.

@time_decorator
def add(x, y=10):
    sleep(2)
    return x + y

is equivalent to

def add(x, y=10):
    sleep(2)
    return x + y

add = time_decorator(add)

The output involving the variable a only happens when time_decorator is called, not when you call the function add.

Understanding excution order in nested functions Python 3.8

1 Answers1