python dela with variable in memory

Question

def choice(n,s=[]):
  a=s
  if n==0:
    string=''
    d=[str(a[len(a)-i-1]) for i in range(len(a))]
    string=string.join(d)
    print(string)
    return string
  if (n-1)%2==0:
    a.append(1)
    n=float(n-1)/2
    return choice(n,a)
  if (n-2)%2==0:
    a.append(2)
    n=float(n-2)/2
    return choice(n,a)
choice(10)#this print 122 ,it's right

choice(10)#but when I run it again ,this print 122122

I dont know what to do

please help me!!!

It's because your list `s` is not = `[]` the 2nd time you run it — Alan Kavanagh, Mar 03 '18 at 02:51

score 1 · Accepted Answer · answered Mar 03 '18 at 02:52

1

Using a mutable list as a default value is bad practice. Simply changing the top of your function like this will fix it.

def choice(n,s=None):
  if s is None:
      s = []

answered Mar 03 '18 at 02:52

G_M

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it works !!!! can you tell me why？ – Kyan Mar 03 '18 at 02:54
@Kyan http://docs.python-guide.org/en/latest/writing/gotchas/#mutable-default-arguments – G_M Mar 03 '18 at 02:55

python dela with variable in memory

1 Answers1