To: Bart From: Ben Bacarisse <ben.use...@bsb.me.uk> Bart <b...@freeuk.com> writes:
> On 23/06/2018 23:25, Ben Bacarisse wrote: >> Bart <b...@freeuk.com> writes: >> >>> On 23/06/2018 21:13, Chris Angelico wrote: >>>> On Sat, Jun 23, 2018 at 10:41 PM, Bart <b...@freeuk.com> wrote: >>> >>>>> (At what point would that happen anyway; if you do this: >>> >>>> NONE of your examples are taking copies of the function. They all are >>>> making REFERENCES to the same function. That is all. >>> >>> This is about your notion that invocations of the same function via >>> different references, should maintain their own versions of the >>> function's 'static' data. >>> >>> Since these references are created via the return g statement here: >>> >>> def f(): >>> def g(): >>> .... >>> return g >>> >>> (say to create function references i and j like this: >>> >>> i = f() >>> j = f() >>> ) >>> >>> I'm assuming that something special must be happening. Otherwise, how >>> does f() know which reference it's being called via? >>> >>> What is different, what extra bit of information is provided when f() >>> is invoked via i() or j()? >> >> f is not being invoked by either i() or j(). i = f() binds i to the >> function returned by f. That's a newly minted function. In languages >> like Python, executing def creates a function. > > Do you mean that if the same 'def' block is re-executed, it will > create a different instance of the function? (Same byte-code, but a > different set of everything else the function uses.) Logically, yes. > Wow. (Just think of all the times you write a function containing a > neat bunch of local functions, every time it's called it has to create > a new function instances for each of those functions, even if they are > not used.) I am surprised that this surprises you, and equally surprised that you seem to think it's going to be in some way grossly inefficient. -- Ben. --- BBBS/Li6 v4.10 Toy-3 * Origin: Prism bbs (1:261/38) -- https://mail.python.org/mailman/listinfo/python-list
