This is nice. One interesting choice is the decision to compute the number of ob...

version_five · on May 2, 2022

> One interesting choice is the decision to compute the number of observed words, N, as the default value of an optional argument, thereby ensuring that it is only computed once

You're saying python computes an expression that's part of a default value when the code defining the function is run? I guess it makes sense now that i say it, I wouldn't have automatically assumed that. That's an interesting one to remember.

darkfirefly · on May 3, 2022

Python's default arguments are evaluated and stored once. This can cause issues, for example the naive "default to []" program:

    >>> def function(b, a=[]):
    ...     a.append(b)
    ...     print(a)
    ... 
    ...     
    >>> function(1)
    [1]
    >>> function(2)
    [1, 2]
    >>> function(3, [4])
    [4, 3]
    >>> function(5)
    [1, 2, 5]

A correct implementation would be:

    >>> def function2(b, a=None):
    ...     if a is None: a = []
    ...     a.append(b)
    ...     print(a)

MisterBiggs · on May 3, 2022

Surprised to see this works for list() as well.

  >>> def func(b, a=list()):
  ...     a.append(b)
  ...     print(a)
  ...
  >>> func(3)
  [3]
  >>> func(3)
  [3, 3]

Honestly surprised this quirk hasn't bit me yet

thrtythreeforty · on May 3, 2022

Yep and that's also the source of the footgun:

def foobar(my_list=[]): my_list.append(42)

A cursory glance suggests this function appends a number to an empty list every time it's run. But actually, the default argument is evaluated once when the function is defined, so calling this function repeatedly will continue to append to that same list.

thatfunkymunki · on May 3, 2022

yeah, this has consequences for constructors too, you don't want to specify default values in the __init__ function signature because of this.