Python 3.8 recursive function saving list across multiple function calls?

Question

I was running into some problems with a project on python 3, and decided to create that smaller version to understand what was happening. Turns out that it just confused me even more. Well, I wrote down that simple recursion based code:

>> Code

1-  def rec(memory=[], x=2):
2-      print(memory, x)
3-      if x > 0:
4-          memory.append(x)
5-          x -= 1
6-          rec(memory=memory, x=x)
7-      else:
8-          memory=[]
9-          return 0
10- 
11- rec()
11- print("First call done.")
12- rec()

And expected something like that:

>> Desired output

[] 2
[2] 1
[2, 1] 0
First call done.
[] 2
[2] 1
[2, 1] 0

But ended up by receiving that:

>> Real output

[] 2
[2] 1
[2, 1] 0
First call done.
[2, 1] 2
[2, 1, 2] 1
[2, 1, 2, 1] 0

Does someone know why the rec() function keeps hold of the memory even when I explicitly tell it to set memory back to an empty list in line 8? And... What do I have to do for it not to save the information across multiple function calls?

you're not setting it back to an empty list like you shown on line 8. that empty list ... `memory = []` is not being called back into the `rec()` function. if it fails the `if` check, it just returns 0. you would need to call `rec()` within that `else` scope then, for that list to be reset to empty. — de_classified, Jun 07 '20 at 22:06

score 1 · Accepted Answer · answered Jun 07 '20 at 21:44

Without explaining the inner workings, you might just do:

def rec(memory = [], x = 2):
    print(memory)
    if x:
        memory.append(x)
        x -= 1
        rec(memory = memory, x = x)
    else:
        del memory[:]
        return

rec()
print("First call done.")
rec()

---

[]
[2]
[2, 1]
First call done.
[]
[2]
[2, 1]

Python 3.8 recursive function saving list across multiple function calls?

1 Answers1