Logs: liberachat/#haskell
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| 2021-06-02 11:36:47 | <ziman> | oh, right, because you'd use the transformer formulation then |
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| 2021-06-02 11:40:30 | <habib> | hey, can anyone tell me if `(a <=< b) . c` is equivalent to `a <=< (b . c)`, and/or how to check the equivalence of different bracketed statements in the general case? |
| 2021-06-02 11:40:48 | <hpc> | :t \a b c -> (a <=< b) . c |
| 2021-06-02 11:40:49 | <lambdabot> | Monad m => (b -> m c) -> (a1 -> m b) -> (a2 -> a1) -> a2 -> m c |
| 2021-06-02 11:40:55 | <hpc> | :t \a b c -> a <=< (b . c) |
| 2021-06-02 11:40:56 | <lambdabot> | Monad m => (b1 -> m c) -> (b2 -> m b1) -> (a -> b2) -> a -> m c |
| 2021-06-02 11:42:04 | <hpc> | you could do it with equational reasoning |
| 2021-06-02 11:42:14 | <merijn> | @check \f g h x -> ((f <=< g) . h) x == (f <=< (g . h)) x |
| 2021-06-02 11:42:15 | <lambdabot> | error: |
| 2021-06-02 11:42:15 | <lambdabot> | • No instance for (Typeable m0) arising from a use of ‘myquickcheck’ • In th... |
| 2021-06-02 11:42:18 | <merijn> | aww |
| 2021-06-02 11:42:23 | <hpc> | expand out the definition of (.) and (<=<), and use the Monad laws to turn them into the same equation |
| 2021-06-02 11:42:40 | <hpc> | or well, same expression |
| 2021-06-02 11:42:56 | <habib> | i guess that's my final resort, just manually expanding it |
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| 2021-06-02 11:44:03 | <habib> | what i really care about is doing `a <=< b . c` without brackets and knowing that it's equivalent to the one i want `a <=< (b . c)`, but if the one i want is also equivalent to the other one, it doesn't matter which way the unbracketed expression resolves to |
| 2021-06-02 11:44:30 | <habib> | in fact, if they're both equivalent, that's happy days, because then it means i don't have to put brackets when i want it the other way as well |
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| 2021-06-02 11:46:36 | <habib> | thanks hpc, i'll just do the expansion tonight |
| 2021-06-02 11:46:46 | <lyxia> | @check \fgh x -> let f,g :: Int -> [Int] ; h :: Int -> Int ; (Fn f, Fn g, Fn h) = fgh in ((f <=< g) . h) x === (f <=< (g . h)) x |
| 2021-06-02 11:46:47 | <lambdabot> | +++ OK, passed 100 tests. |
| 2021-06-02 11:47:55 | <hpc> | jeez, i need to properly learn quickcheck |
| 2021-06-02 11:48:15 | <habib> | oh nice |
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| 2021-06-02 11:49:33 | <habib> | that's confidence enough for me with what i'm doing if i do a manual test and things still work the same |
| 2021-06-02 11:49:39 | <habib> | thanks lyxia |
| 2021-06-02 11:50:09 | <habib> | i gotta get back to work now ha; i'll use this tonight |
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| 2021-06-02 12:21:29 | <jacks2> | hi. where can I find textual source that is in the gif on this page? http://hackage.haskell.org/package/concurrent-output |
| 2021-06-02 12:21:32 | <jacks2> | I tried google but no luck |
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| 2021-06-02 12:23:21 | <Taneb> | jacks2: https://git.joeyh.name/index.cgi/concurrent-output.git/tree/demo2.hs |
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| 2021-06-02 12:23:52 | <jacks2> | thanks! |
| 2021-06-02 12:23:55 | <Taneb> | I went to the website linked to as the source repo on Hackage, and saw there was a demo2.hs file |
| 2021-06-02 12:25:12 | <jacks2> | yeah, I missed it |
| 2021-06-02 12:25:17 | <jacks2> | pretty nifty library. |
| 2021-06-02 12:26:02 | <jacks2> | demo works flawlessly on linux, would be nice if it worked on other OSes as well. I'll have to test |
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| 2021-06-02 12:31:05 | <Guest5> | Hi, I have a strange Haskell problem |
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| 2021-06-02 12:31:36 | <Guest5> | I have a list `[c, b, a]`, I then call `reverse` on the list and it gives back `[b, a, c]` |
| 2021-06-02 12:31:51 | <Guest5> | How is that possible? |
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| 2021-06-02 12:32:09 | <jacks2> | > reverse [1, 2, 3] |
| 2021-06-02 12:32:11 | <lambdabot> | [3,2,1] |
| 2021-06-02 12:32:11 | <merijn> | > reverse [ 3, 2, 1] |
| 2021-06-02 12:32:13 | <lambdabot> | [1,2,3] |
| 2021-06-02 12:32:43 | int-e | can think of dirty runST/unsafeInterleaveST tricks to do that kind of thing |
| 2021-06-02 12:32:44 | <merijn> | Pretty sure you're calling the wrong reverse function |
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