Code undoes itself before returning value-CodePudding

from time import sleep
def refit(i, n, c=[]):
 sleep(1)
 print(c)
 if i[:n] != '': refit(i[n:],n,c [i[:n]])
 sleep(1)
 print(c)
 return c

Output when using arguments "Helloo" and 2:

[]
['He']
['He', 'll']
['He', 'll', 'oo']
['He', 'll', 'oo']
['He', 'll']
['He']
[]
[]

The Array ['He', 'll', 'oo'] should be printed and returned, but the code is undoing itself. What is happening? I am so confused.

CodePudding user response：

Did you mean to save the return value from recursive calls to refit()?

def refit(i, n, c=[]):
    if i[:n] != '':
        c = refit(i[n:], n, c [i[:n]])
    return c

c = refit('Helloo',2)
print(c)

Output as requested

CodePudding user response：

Within a given call to refit, you're not changing c, so c has the same value in the second print() that it does in the first print(). All the recursive calls happen in between those two print calls.

On the first call, c is []:

# i = 'Helloo', n = 2, c = []
print(c)  # prints "[]"
if i[:n] != '':
    refit(i[n:],n,c [i[:n]])  # prints a bunch of other stuff
print(c)  # prints "[]"
return c  # returns []

On the second call to refit (indicated above by # prints a bunch of other stuff) we do:

# i = 'lloo', n = 2, c = ['He']
print(c)  # prints "['He']"
if i[:n] != '':
    refit(i[n:],n,c [i[:n]])  # prints even more stuff
print(c)  # prints "['He']"
return c  # returns ["He"]

Our output so far looks like:

[]
['He']
(more stuff will go here)
['He']
[]

And on and on -- each new recursive call adds another pair of print statements in the middle of the previous call. And each call returns its original value of c, rather than the new value it got from the recursive call it just made to refit.

If you want this function to return the result from the innermost call, then it simply needs to return the result of calling refit:

def refit(i, n, c=[]):
    return refit(i[n:], n, c   [i[:n]]) if i[:n] else c

print(refit("Helloo", 2))  # ['He', 'll', 'oo']