CODEWITHSUNDEEP
python
cgoma
cgoma

Reputation: 49

Issue with passing list as argument to python function

def foo(val, arr=[]):
    arr.append(val)
    return arr

print(foo(1)) ---> outputs [1]
print(foo(2)) ---> outputs [1, 2]

Above happens because the default value (an empty list) was evaluated once, when the function was compiled, then re-used on every call to the function. If this is the case then I am not able to understand output of below python code

def f(x,l=[]):
    for i in range(x):
        l.append(i*i)
    print(l) 

f(2)
f(3,[3,2,1])
f(3)
#This outputs below
[0, 1]
[3, 2, 1, 0, 1, 4]
[0, 1, 0, 1, 4]

Here in second call f(), list has been changed to [3, 2, 1, 0, 1, 4] but in the third call to f() uses list created in first call f() i.e. [0, 1]

I am unable to understand how third call is not using list crated during 2nd f() call.

Upvotes: 1

Views: 1200

Answers (1)

Chris Charles
Chris Charles

Reputation: 4446

First gotcha in the list http://docs.python-guide.org/en/latest/writing/gotchas/

The list is only created once, so if you modify it then that modified version is used next time.

In the second call f(3, [3,2,1]); it isn't using the default list, it's using the one supplied by you. So when you call a third time you're using the default list which was only modified by the first call.

as an example, looking up the memory addess for each object using id:

def foo(a=[]):
    print id(a)

foo()   #140378224732136
foo([]) #140378224687944
foo()   #140378224732136

Upvotes: 3

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