params=(): a new way to work around python's static default args

mutable_default_args.py

# coding:utf-8
# Which one of these approaches is best?

def append_bar(params=[]):
    '''
    Nope. This has a well-known problem: default args are created at definition,
    not at call, so the globally-scoped `[]` value ends up getting reused
    whenever we don't specify `params`:

    >>> append_bar()
    ['bar']
    >>> append_bar()
    ['bar', 'bar']
    '''
    params.append("bar")
    return params


def append_bar2(params=None):
    '''
    The standard work around is this, and works as expected:

    >>> append_bar2()
    ['bar']
    >>> append_bar2()
    ['bar']
    '''
    params = params if params is not None else []
    params.append("bar")
    return params

def append_bar3(params=()):
    '''
    My new approach.

        - 👍 clearer that params is expecting a seq
        - 👍 concise
        - 👎 extra copy

    >>> append_bar3()
    ['bar']
    >>> append_bar3()
    ['bar']
    '''
    params = list(params)
    params.append("bar")
    return params

def subst(params=()):
    '''
    And in the situation where we can guarantee we only read params:
        - 👍 we can avoid munging it altogether

    >>> subst(params=('bar', 'baz'))
    'foo? bar;foo? baz;'
    '''
    return ('foo? %s;' * len(params)) % params