我对Python3.6中默认参数值的困惑

Question

Here are two pieces of codes which were under the standard of python3.6. And they are the examples in the docs of python3.6(tutorial, page25). The first is:

def f(a, L=[]):
    L.append(a)
    return L
print(f(1))
print(f(2))
print(f(3))

the result:

[1]
[1, 2]
[1, 2, 3]

the second:

def f(a, L = None):
    if L is None:
        L = []
    L.append(a)
    return L
print(f(1))
print(f(2))
print(f(3))

the result:

[1]
[2]
[3]

So, in the second piece of code, i am confused that after print(f(1)) was executed, print(f(2)) would pass a = 2 and L=[1] to the f() , but why f() didn't get the L=[1] ? If L = None in the second piece of code defines the L to None every time when the f() was called, but why L = [] in the first piece of code don't define L to []

Answer 1

Those two examples show how default arguments work behind the scenes: the first one demostrates that default arguments 'live' inside the function definition. Meaning: that the value for L in the first function will only ever be reset if you overwrite the whole function with a def section.

The Same is true for the second implementation BUT since it's None: you have to initialize it while the function body is executed. This leads to a fresh list every time the function is called.

This behaviour can be confusing and lead to strange results which is why i heard from most sources that it is best to avoid the first option and work with None default args.

Hope i could clear things up a bit.