Python: Default list in function -

from summerfield's programming in python3:

it says follows: when default values given, created @ time def statement executed, not when function called. question following example:

def append_if_even(x, lst =none):     lst = [] if lst none else lst     if x % 2 ==0:         lst.append(x)     return lst 

as first time definition executed, lst point none after function call append_if_even(2),

• shouldn't lst point [2], since after lst.append(x) lst not point none anymore ?

• why next execution still make lst point none?

• what happen inside function call append_if_even(2)?

shouldn't lst point [2], since after lst.append(x) lst not point none anymore? why next execution still make lst point none?

that prevent using lst=none, lst = [] if lst none else lst construction. while default arguments function evaluated once @ compile time, code within function evaluated each time function executed. each time execute function without passing value lst, start default value of none , replaced new empty list when first line of function executed.

if instead define function this:

def append_if_even(x, lst=[]):     if x % 2 ==0:         lst.append(x)     return lst 

then act describe. default value lst same list (initially empty) every run of function, , each number passed function added 1 growing list.

for more information, see "least astonishment" , mutable default argument.