Comment by codesnik

5 days ago

you could just treat argument initialization as an executable expression which is called every time you call a function. If you have a=[], then it's a new [] every time. If a=MYLIST then it's a reference to the same MYLIST. Simple. And most sane languages do it this way, I really don't know why python has (and maintain) this quirk.

4 comments

codesnik

1718627440 5 days ago

What are the semantics of the following:

    b = ComplexObject (...)
    # do things with b

    def foo (self, arg=b):
        # use b

    return foo

Should it create a copy of b every time the function is invoked? If you want that right now, you can just call b.copy (), when you always create that copy, then you can not implement the current choice.

Should the semantic of this be any different? :

    def foo (self, arg=ComplexObject (...)):

Now imagine a:

    ComplexObject = list

codesnik 5 days ago
I wonder, why that kind of ambiguity or complexity even comes to your mind at all. Just because python is weird?
def foo(self, arg=expression):
could, and should work as if it was written like this (pseudocode)
def foo(self, arg?): if is_not_given(arg): arg=expression
if "expression" is a literal or a constructor, it'd be called right there and produce new object, if "expression" is a reference to an object in outer scope, it'd be still the same object.
it's a simple code transformation, very, very predictable behavior, and most languages with closures and default values for arguments do it this way. Except python.
- 1718627440 5 days ago
  
  What you want is for an assignment in a function definition to be a lambda.
  def foo (self, arg=lambda : expression):
  Assignment of unevaluated expressions is not a thing yet in Python and would be really surprising. If you really want that, that is what you get with a lambda.
  > most languages with closures and default values for arguments do it this way.
  Do these also evaluate function definitions at runtime?
  
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