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I don't actually understand what is happening here.

You see the code below, I want to know what does the 'return' statement actually returns and to where? Is [3,4,2,5] the value of calc(list[1:]) and does it change or not? def calc(list): if len(list)==0: return 0 else: return list[0]**2 + calc(list[1:]) list = [1, 3, 4, 2, 5] x = calc(list) print(x) o

1st Nov 2021, 7:32 PM
Mas'ud Saleh
Mas'ud Saleh - avatar
1 Odpowiedź
+ 1
If you just call calc like this list = [1, 2, 3] calc(list) the return goes to nowhere. You can save it into a variable like in your postet code. x = calc(list) clalc returns 0 if the list is empty. If not this happens: 1. It takes the first value from list and squares it. 2. Takes a the rest from the list and calls calc with it again. This happens so long till there is no rest, so the given list is empty. If you input was [1,2,3] the result is 14. calc([1,2,3]) # 14 1*1 + calc([2,3]) # 1 + 13 2*2 + calc([3]) # 4 + 9 3*3 + calc([]) # 9 0
1st Nov 2021, 8:27 PM
Stefanoo
Stefanoo - avatar