[klyrs]

>

   # Are these two lines actually faster than the
   # interpreter creating defaults at call time?

You're proposing that the interpreter add a check for every default parameter in every function signature; that it should optionally fire off arbitrary code for each and every one. And when you consider that high-performance Python involves writing C extensions, your proposal would be to move that check out of the compiled code and into the slow interpreted space is, yes, a major performance hit.

[kerkeslager]

> You're proposing that the interpreter add a check for every default parameter in every function signature

No, that's not what I'm proposing. Why would it check anything? Just evaluate the given default expression at call time. If you don't want the overhead of an expression, don't put a default.

You can also do defaults like:

    LIST_OF_X = []

    def foo(xs = LIST_OF_X):
        ...

...if you want the other behavior. This does add a variable lookup (oh no!).

[klyrs]

Your list there is mutable. Isn't that what you meant to solve with this?

[kerkeslager]

Ugh. You're just ignoring my entire post except the one part where you (wrongly) think you can correct me?

The code I posted is showing how you can explicitly get the mutable behavior if you want it when the default expressions are evaluated at call time.

That is to say, if you evaluate default expressions at call time, you can get either behavior by being explicit with minimal loss in performance:

    def foo(xs = []):
        xs.append(1)
        print(xs) # always prints [1]

    DEFAULT = []
    def bar(xs = DEFAULT):
        xs.append(1)
        print(xs) # prints [1], [1,1], [1,1,1], etc.

To reiterate, the above code is what would happen if default expressions were evaluated at function call time.

You're clearly not as knowledgeable as you think you are on this and you're just cherry picking things you don't understand to feel smart.