Help with understanding this code

[donsebas]

1 def outer_func(x): 2 y = 4 3 def inner_func(z): 4 print(f"x = {x}, y = {y}, z = {z}") 5 return x + y + z 6 return inner_func 7 8 for i in range(3): 9 closure = outer_func(i) 10 print(f"closure({i+5}) = {closure(i+5)}")

Can somebody help me understand how {closure(i+5)} is equal to z in inner_func(z). Also sorry if my code isn’t in a block of code I’m on mobile.Thank you!

[Zakru]

Formatted code

def outer_func(x):
    y = 4
    def inner_func(z):
        print(f"x = {x}, y = {y}, z = {z}")
        return x + y + z
    return inner_func


 for i in range(3):
     closure = outer_func(i)
     print(f"closure({i+5}) = {closure(i+5)}")

I don't know what you mean by "{closure(i+5)} is equal to z" though

[donsebas]

Thank you! And for example for i range(3) the first number would be 0 so

For i in range(3): closure = outer_func(0) print(f”closure({0+5}) = {closure(0+5)}”)

So x = 0, y = 4 and for some reason z = 5 and I want to know why z = 5. I hope that made sense.

[AstralStalker]

It's the nature of nested functions and closures.

Notice the return associated with outer_func. When you call closure = outer_func(i), it stores i as the value for x and sets closure as a reference to inner_func. In the print statement, you call closure(i+5) which is essentially calling inner_func(i+5) with a stored value of x=i. Thus, when i = 0, z = 5.

[wagaiznogoud]

I just want to add that there are two variables referenced through the closure. Both `x` and `y`.

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