为什么 Python 3 中函数中的初始化字典参数是不可变的？

Question

In the following defined function,

def a_function(x, dictionary = dict()):
    if not dictionary:
        print('Empty Dictionary')
        dictionary['A'] = 1
        dictionary['B'] = 2
    return dictionary.get(x)

it takes in 'A' or 'B' and returns 1 or 2 . The dictionary argument is supposed to be initialized each time, but when I run it, it is immutable, in that once created, it doesn't start again. 'Empty Dictionary' is run only once. Why?

>> a_function('A')
Empty Dictionary
1
>> a_function('B')
2

Answer 1

Python docs states that:

Default parameter values are evaluated from left to right when the function definition is executed.

This means that the expression is evaluated once, when the function is defined, and that the same “pre-computed” value is used for each call .

For example:

>>> import datetime
>>> def f(foo=datetime.datetime.now()):
...     print(foo)

>>> f()
2020-01-21 05:21:35.084471
>>> f()
2020-01-21 05:21:35.084471