until, laziness, and performance
george.colpitts at gmail.com
Sun Mar 27 22:52:27 UTC 2016
Got it! Thanks!!
On Sun, Mar 27, 2016 at 7:44 PM Dan Doel <dan.doel at gmail.com> wrote:
> On Sun, Mar 27, 2016 at 4:42 PM, George Colpitts
> <george.colpitts at gmail.com> wrote:
> > I still don't understand why
> > as' i s
> > | i >= n = s
> > | otherwise = as' (i + 2) (s + (-1) / i + 1 / (i + 1))
> > is not lazy in its second arg as it only has to evaluate it if i >= n.
> > seems exactly analogous to or, ||, which only needs to evaluate its
> > arg if its first arg is False. || is definitely lazy in its second
> > Can you explain why as' is strict in its second arg, s? Is it due to the
> > following from,
> > https://ghc.haskell.org/trac/ghc/wiki/Commentary/Compiler/Demand ?
> The simplest way to see why as' is strict in s is to use the actual
> definition of strictness, and not think about operational semantics.
> The definition is that a function f is strict if `f undefined =
> undefined`. So, what is `as' i undefined`?
> If `i >= n` then `as' i undefined = undefined`
> if not `i >= n` then `as' i undefined = as' (i+2) undefined`
> The second case holds because Double arithmetic is strict, so the
> complex expression will be undefined because s is.
> From these cases we can reason that the result will always be
> undefined. Either the value of i will grow such that the first case
> succeeds, or if that will never happen (if n is NaN for instance), the
> loop will continue forever, in which case the result is equivalent to
> undefined semantically.
> Because of this denotational analysis, we are at liberty to make the
> operation of as' such that it always evaluates s immediately, because
> it won't change the answer. And that is the optimization that saves
> By contrast, my alternate function:
> as' i s
> | i >= n = (i, s)
> | otherwise = as' (i+2) (...)
> is easily not strict in s, because:
> as' i undefined = (i, undefined)
> when `i >= n` is True, and (i, undefined) is not undefined. So it is
> invalid to perform the same optimization, because it would change the
> answer on some inputs. We happen to know that all the places the
> function is used, those two answers would be equivalent (because we
> just project out the second item), but that's somewhat difficult,
> non-local reasoning.
> Note that GHC does do some optimization. The loop in until is (in this
> case) a loop on a pair, and it figures out that the loop is strict in
> the pair, and eliminates the construction of the pair (by passing two
> separate arguments, and returning an unboxed pair from the loop).
> Optimizing the pieces of the pair would have to be a sort of
> speculative operation (without doing a global analysis), where 2^k
> different versions of the loop are generated (where k is the number of
> non-strict pieces), and the call sites are optimized to call the best
> of those versions for them. But 2^k isn't a good looking number for
> this sort of thing.
> -- Dan
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