Python, loops and closures

I'm a fairly experienced C/C++ (and to some degree, Java) programmer. I'm learning python, but I'm baffled at some strange (for my backgroung) behaviors of the language.

I'm learning about nested function and closures (reading "Learning Python", that seems a really good source for me).

I understand that if I nest a def inside a for loop when I call the created function, it looks up the last value of the captured loop variable (as it captures by reference, as a C++ programmer would put it)

funcs = []
for i in range(4):
    def f():
        print(i)
    funcs.append(f)

and running the program the result is

>>> for f in funcs:
      f()


3
3
3
3

Now, I was wrapping my head around this when I stumbled upon this (what to me seems) an inconsistency: if I do

for i in range(4):
  funcs[i]()


0
1
2
3

more baffling, if I do

>>> i = 2
>>> funcs[i]()

2

and now, all functions in list returns 2:

for f in funcs:
  f()


2
2
2
2

there must be some scope related question that I can't grasp

Solution

First, this creates a list of four functions.

funcs = []
for i in range(4):
  def f():
    print(i)
  funcs.append(f)

Each of these functions looks up the value of i and then prints it.

This loops through the list of function and calls each one:

>>> for f in funcs:
      f()

As stated above, these functions look up i, which is 3 right now due to the for i in range(4) loop that completed earlier, so you get four printouts of 3.

Now you loop again, using i as the loop variable:

for i in range(4):
  funcs[i]()


0
1
2
3

The first time through the loop, i is 0, so when the function looks up i, it gets 0, and then prints that. Then it changes to 1, then 2, then 3.

The following code simply changes i in yet another way, and calls a function:

>>> i = 2
>>> funcs[i]()

2

You could've called any of those functions and they still would've printed 2, because that's the value of i now. You're just getting lost because you looped over range(4) to create these functions, then you looped over range(4) to index the list of functions, and you keep reusing i, and then you reassign i and also use it to index the list.

If you want each function's printed value of i to be fixed at what it was when you defined the function, the easiest way to do that is with a default argument, as those are evaluated when the function is defined rather than when it's called:

funcs = []
for i in range(4):
  def f(num=i):
    print(num)
  funcs.append(f)