🔒 This Output Prediction Question !

[vital remnant]

Hello, Could someone pls explain the logic of this code and its output???

[tall whale]

@vital remnant

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[lavish widget]

funcs = []
this is a list, it is currently empty however

for i in range(3): # counts from 0 - 2 (0, 1, 2) -> That's 3 numbers!
    def f():
        return i # returns i, which is a number
    funcs.append(f) # add the function 'f' to the list
````funcs` should now have 3 `f`s in it

```py
for fn in funcs: # Iterate through the list, i.e go through each element in the list
  print(fn()) # 'fn()' calls the function, then we print what the function returns

[vital remnant]

ok but why does it print 2 ?

[lavish widget]

imagine this,

for fn in funcs: # Iterate through the list, i.e go through each element in the list
  print(fn()) # 'fn()' calls the function, then we print what the function returns

is the same as

fn()
fn()
fn()

And since fn does return i and previously we looped all the way until 2, it will always output 2

TLDR; we aren't storing the value of i, we are storing the variable, and that variable has a value of 2 at the end of the for i in range(3) loop

[light shell]

wait I got a bit confused with appending f instead of f()

[light shell]

your appending the function object to the list. As you are appending outside the function the maximum value of i is appended but you didn't call the function. the function object that returns 2 is appended. In the second loop when you iterate through the list. you are calling the function object which is returning the value 2 as the other values through the loop are not stored which is why you see only twos instead of 0, 1, 2

if you wanted 0,1,2 then you would append i inside the function

[tall whale]

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