+ 4
[Solved] Why don't all list elements get zeroed in this code?
18 Antworten
+ 4
no: primitives are passed by values, while objects are passed by references ^^
that's why only the first item is set to zero: the list id at the initial recursion call remain the same, but each other call pass the sliced list (wich is a copy) ;)
+ 3
in 'Recurse' function you're trying to write in argument and you return nothing: a remain holding the same list/memory adress.
only first item of list is set to zero, because the fact is that at each recursion steps, you're passing a copy of the argument list subset (slicing): you could print(id(a)) before Recursion call inside Recursion ;)
+ 2
visph so I guess, the following is the way to achieve my desired outcome then.
https://code.sololearn.com/cD8D3a49UxD6/?ref=app
+ 2
to recursively set to zero all items of a list, rather use something like:
def Recurse(a,i=0):
if i != len(a):
a[i] = 0
Recurse(a,i+1)
however, a basic loop would be more efficient and not limited ^^
+ 2
yes, your solution is valid too, but you doesn't need to return the Recursion call results ;)
+ 2
with this one, you print another list inside 'recurs' function, but global list 'a' after recursion is [2,3,4] ;)
+ 2
Lol, i am not after a completely new list but the original one modified. But thanks for trying.
+ 1
The list ids are the same before and after recursion.
+ 1
visph I see. It's the slicing that causes the copy. I think I get it now.
+ 1
visph yes I know that a simple loop is more efficient. This was just a study of recursion.
+ 1
visph merçi beaucoup.
+ 1
avec plaisir :D
+ 1
solvermad
but if you do so, the array will be empty at end of recursion ^^
+ 1
jeez, i just cant win
+ 1
its a good code, trying to solve it i learned about the nonlocal keyword & scoping, but i coulnt make them work for your problem. still its a nice code Sonic
0
inside the else statement you could do...
a.remove(a[0])
print(id(a))
Recurse(a)
0
def recurs(a):
a.pop()
a = [0] + a
if a == [0,0,0,0]:
return print(a)
else:
(recurs(a))
a = [1,2,3,4]
recurs(a)