2025/01/08

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2025-01-08 11:23:14 +0100 <tomsmeding> if you want to read manually from a MutableByteArray, https://hackage.haskell.org/package/base-4.19.0.0/docs/GHC-Exts.html#v:readInt64Array-35-
2025-01-08 11:22:38 +0100 <tomsmeding> are you sure you're not looking for Data.Vector.Storable.Mutable?
2025-01-08 11:21:43 +0100 <tomsmeding> one containing the fst components, the other containing the snd components
2025-01-08 11:21:28 +0100 <tomsmeding> an `MVector s (Int, Int)` in the end consists of _two_ `MutableByteArray s`es
2025-01-08 11:20:29 +0100 <tomsmeding> and also, the 'a' here is a boxed type
2025-01-08 11:20:10 +0100 <tomsmeding> but unboxed types in an unboxed vector (as opposed to a _storable_ vector) are stored in struct-of-arrays format
2025-01-08 11:19:31 +0100 <tomsmeding> recall my earlier comments on the ghc unboxer
2025-01-08 11:19:20 +0100 <tomsmeding> are you sure you need that for performance?
2025-01-08 11:17:34 +0100Guest17(~Guest17@91.243.81.204)
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2025-01-08 11:15:55 +0100tnt2tnt1
2025-01-08 11:13:37 +0100tnt2(~Thunderbi@user/tnt1) tnt1
2025-01-08 11:13:34 +0100tnt1(~Thunderbi@user/tnt1) (Ping timeout: 272 seconds)
2025-01-08 11:07:16 +0100 <haskellbridge> <magic_rb> is there no way to directly read from a Unboxed Mutable Vector using Int# as the index and read the undoxed type directly? https://hackage.haskell.org/package/vector-0.13.2.0/docs/Data-Vector-Unboxed-Mutable.html i dont see anything in the module
2025-01-08 11:06:06 +0100tomsmeding(~tomsmedin@2a01:4f8:c0c:5e5e::2) tomsmeding
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2025-01-08 11:01:01 +0100merijn(~merijn@77.242.116.146) merijn
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2025-01-08 10:55:22 +0100CiaoSen(~Jura@2a05:5800:213:8800:ca4b:d6ff:fec1:99da) CiaoSen
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2025-01-08 10:43:31 +0100 <ncf> monochrom: what relation do you use for b? i'd think a function Bool → b is enough
2025-01-08 10:41:21 +0100jespada(~jespada@2800:a4:f9:a300:c8c1:20a5:1a94:f1) jespada
2025-01-08 10:37:49 +0100merijn(~merijn@77.242.116.146) (Ping timeout: 244 seconds)
2025-01-08 10:37:16 +0100m_mari-estel
2025-01-08 10:36:48 +0100mari-estelm_
2025-01-08 10:36:44 +0100tzh(~tzh@c-76-115-131-146.hsd1.or.comcast.net) (Quit: zzz)
2025-01-08 10:34:56 +0100ash3en(~Thunderbi@2a03:7846:b6eb:101:93ac:a90a:da67:f207) ash3en
2025-01-08 10:33:28 +0100eL_Bart0(eL_Bart0@dietunichtguten.org)
2025-01-08 10:31:56 +0100lxsameer(~lxsameer@Serene/lxsameer) lxsameer
2025-01-08 10:28:00 +0100 <lambdabot> (forall x. g . h x = k (f x) . g) => g . fold h y = fold k (g y) . $map f
2025-01-08 10:28:00 +0100 <monochrom> @free fold :: (x -> y -> y) -> y -> [x] -> y
2025-01-08 10:22:49 +0100Smiles(uid551636@id-551636.lymington.irccloud.com) Smiles
2025-01-08 10:18:14 +0100l_k(~student@85.172.76.134) (Read error: Connection reset by peer)
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2025-01-08 10:10:34 +0100alist(~alist@108-224-153-186.lightspeed.cicril.sbcglobal.net)
2025-01-08 10:05:14 +0100 <monochrom> But the paper doesn't contain such advanced examples. All examples are basically natural transformations like "oh rev is a natural transformation". Those don't need arbitrary relations, just functions.
2025-01-08 10:03:59 +0100 <monochrom> But if you replace Maybe by the System F encoding, i.e., prove that e :: forall a. forall b. b -> ((a -> a) -> b) -> b has only two possibilities, that will require you to use a non-function relation for b.
2025-01-08 10:03:02 +0100lortabac(~lortabac@2a01:e0a:541:b8f0:55ab:e185:7f81:54a4)
2025-01-08 10:02:00 +0100 <monochrom> Err, that one still works with a function relation.
2025-01-08 10:01:00 +0100 <monochrom> Sometimes you really want to use a non-function relation, e.g., when proving that e :: forall a. Maybe (a->a) can only be Nothing or Just id.
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2025-01-08 09:56:30 +0100machinedgod(~machinedg@d108-173-18-100.abhsia.telus.net) machinedgod