How returns behave in a function with multiple returns?

[khasbay]

The code below returns 8. Why does not it return 2 instead? I do not understand how return behaves in the code below. what is the rule? Thanks for your help.. khasbay

def f(x):
    if x==0:
        return 2
    else:
        return x+f(x-1)

print(f(3))

deanhystad write May-19-2024, 01:54 AM:
Please post all code, output and errors (it it's entirety) between their respective tags. Refer to BBCode help topic on how to post. Use the "Preview Post" button to make sure the code is presented as you expect before hitting the "Post Reply/Thread" button.

[deanhystad]

Follow the function calls to compute the value for f(3).

You call your function f(3).

In the function, 3 != 0, so the function returns 3 + f(2).

We call f(2). 2 != 0, so the function returns 2 + f(1).

We call f(1). 1 != 0, so the function returns 1 + f(0).

We call f(0). 0 == 0, so the function returns 2

We have these values for f()
f(0) = 2
f(1) = 1 + f(0) = 3
f(2) = 2 + f(1) = 5
f(3) = 3 + f(2) = 8

Recursion is when a function calls itself. the value of f(3) is slightly confusing not because it has two return statements, but because it uses recursion. This is the same function with one return statement.

def f(x):
    return 2 if x == 0 else x + f(x-1)

print(f(3))