Python variable weird issue

Question

sorry i really do not know how to describe this issue accurately in the title

I define a function like this

def f(v,l=[]):
    l.append(v)
    return l

In my understanding, the output should be like this:

the first call should return [0]

the second call should return [1]

the third call should return [2]

But.. here is the real output

>>> f(0)
[0]
>>> f(1)
[0, 1]
>>> f(2)
[0, 1, 2]

don't use `l=[]` because Python creates this `[]` only once when script is loaded. — furas, Nov 14 '16 at 05:32
To be more precise, default arguments are evaluated when the function definition is executed, not when the function is called. — PM 2Ring, Nov 14 '16 at 05:42

furas · Answer 1 · 2016-11-14T05:44:38.163

4

You have to do

def f(v, l=None):
    if l is None:
        l = []
    l.append(v)
    return l

because in l=[] this list [] is created only once when script is loaded.

More precise (as @PM2Ring said) it is created when the function definition is executed, not when the function is called.

edited Nov 14 '16 at 05:44

answered Nov 14 '16 at 05:34

furas

1 Answers1