Re: OO versus RDB
Date: Sun, 9 Jul 2006 23:25:20 -0600
[Note that I'm just picking out interesting things to talk about. I could care less about the thread to this point, nor the particular definitions in use; and please feel equally free to ignore me.]
Marshall <marshall.spight_at_gmail.com> wrote:
> > Definitions are important; that's why we
> > have so many bunfights in these newsgroups about them.
> No. *Agreeing* on definitions is important; the specific definitions
> themselves are unimportant.
There is one commonly missed sense in which definitions are quite important. Definitions that occupy common words (linguistic real estate) and express concepts that are not fundamentally interesting are harmful.
And there, indeed, is quite an uninteresting word. (Granted, it's not occupying a particularly attractive piece of linguistic real estate, either!)
> Please remember: my only points were
> 1) A parameterized abstraction (by whatever name) that
> contains writes to variables but no reads has its return
> value determined exclusively by its inputs, despite the
> 2) This is useful to know
No outright disagreement, but I'm not sure what standard of "useful" is being used for #2, or in what context. Useful is a matter of degree. This harkens somewhat back to your question about which "side effects" are important, and which are not. If the write to the variable affects whether the program produces correct results (i.e., is "important"), then it complicates the model similarly to any other write operation. The model of the language now needs to contain some concept of time (or, equivalently, state). That's a big deal. If that's not true, then of course it doesn't matter, and you just need a way to let your tools know that (and if the "tool" is your brain for the current task, than that's presumably trivial).
So (I'm working this out as I go; please excuse any errors, and please point them out...) in the interesting case, the time aspect of the model is most flexibly expressed as a set of statements of the forms:
P(a,b,C): if expression b is evaluated , then a must be evaluated
before b with no intervening expressions in C.
One can write rules for the time dependencies introduced.
For general-case functional subexpressions (or functions under applicative order), P(a,b,0) [0 is the empty set] if a is a subexpression (argument) of b.
And for shared state, P(a,b,C) if:
(1) a writes state s(a) to a shared memory slot (2) b reads from the shared memory slot and expects to find s(a,E). (3) C contains all C[i] that write s(C[i]), where it is not provable that s(C[i]) = s(a).
Knowing that the function (called f from here on) doesn't depend on anything except its parameters means that it will never occur in the "b" position due to shared state; which means that if P(a,f,C) then this is a time constraint due to applicative order, so C is the empty set. However, note that f can occur in a P(f,b,C), and also that it may occur in some set C for which P(a,b,C). Sure, that's a useful thing to know. If all functions worked that way, then in fact you'd be looking at the same situation as applicative order again. If only some functions work that way, then what the statements that can be made are relatively weak, since two out of three ways for f to exist in a complex ordering constraint to exist are still applicable. On the other hand, if f neither depends upon nor modifies any "important" state, then we are guaranteed that f only occurs in constraints for which C is the empty set, which leaves a lot more room for manipulation.
Admittedly, the above is a horrible, inexcusable simplification, because it fails to account for functions that both depend upon AND modify shared state. When they get added in, your mutation-only function will start to look a lot more appealing. The s(a) expressions above will become s(a, ENV), and you'll be able to assert that for mutation-only functions, s will be independent of the ENV parameter, and that will help avoid a combinatorial explosion of possible program behaviors. However, I lack the clarity of thought this late at night to make it formal.
I'm a little out of context here, so you may perfectly well understand this... but I suspect any rational objections you get to such a statement come from the concern that the results of saying that a specific function (as opposed to all functions) is mutation-only are considerably weaker in the order analysis above than what you'd get for a function that neither depends upon nor modifies global state.
-- Chris Smith - Lead Software Developer / Technical Trainer MindIQ CorporationReceived on Mon Jul 10 2006 - 07:25:20 CEST