February 2015 - Haskell-Cafe

A Question about IO
by Zongzhe Yuan 28 Feb '15

28 Feb '15

Hi i have a question about IO monad. Sometimes IO do something and return something, i wonder if the return type, for example is IO Int means it will return an int, could i purely fetch the int? but i fail, i cannot construct a function, i use do notation but i found i must use something to pass the int to another monad. But i could write a function has the type of Maybe a -> a and it is easily. Maybe monad and IO monad both instance monad but why i can’t do that? Zongzhe Yuan This message and any attachment are intended solely for the addressee and may contain confidential information. If you have received this message in error, please send it back to me, and immediately delete it. Please do not use, copy or disclose the information contained in this message or in any attachment. Any views or opinions expressed by the author of this email do not necessarily reflect the views of the University of Nottingham. This message has been checked for viruses but the contents of an attachment may still contain software viruses which could damage your computer system, you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation.

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Parallel numeric class hierarchy
by Clinton Mead 28 Feb '15

28 Feb '15

Let's say for arguments sake I want to make a new finer grained numeric class hierarchy. But, I also want to have the following occur: (1) Existing modules which have defined data types which have Num instances can be used with functions in my new hierarchy (2) New data types defined in the new hierarchy automatically get Num instances if appropriate so they can be used in existing modules. I know this is somewhat of a vague question but roughly how would I go about this?

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ANN: diagrams-hsqml - a Qt5, QtQuick Canvas backend
by Marcin Mrotek 28 Feb '15

28 Feb '15

Hello, I wrote a Diagrams backend that paints on QtQuick (Qt5) canvases through HsQML. It's somewhat half-baked at this point and I haven't fully tested it, but it's good enough for my puroposes (rendering charts through Chart-diagrams) and I have more urgent things to do right now, so I decided to release it as it is. http://hackage.haskell.org/package/diagrams-hsqml https://github.com/marcinmrotek/diagrams-hsqml https://github.com/marcinmrotek/diagrams-hsqml/issues There's one bug I already know of but have no idea how to fix; namely when rendering charts from Chart-diagrams all text is fully filled, with a properly rendered black tex laid on top of it. This can be worked around by setting the font colour to transparent. Bug reports and help in improving the backend by all means welcome. Best regards, Marcin Mrotek

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Bounded STM Queues with load shedding
by Alex Hammel 28 Feb '15

28 Feb '15

Forwarding from Haskell-Beginners: Hi list, It would be handy for me to have a bounded STM Queue along the lines of TBQueue with the difference that when the queue is full, a call to writeTBQueue is a non-blocking no-op. This is in the context of an asynchronous logging system. If the load (somehow) gets heavy enough that the log message queue is taking up a lot of memory, It's better to lose a few log messages than to have processes block until the logger catches up. Are there any off-the-shelf solutions? Cheers, Alex

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Hackage and Free Software
by fr33domlover 28 Feb '15

28 Feb '15

Hello haskellers! I would like to make a proposal regarding the license of software in Hackage. One of the major parts of the Haskell community infrastructure in the package database server Hackage. As far as I know - please correct me if I'm wrong - Hackage makes no restriction on the license of packages upload to it. But as a community working in cooperation to make good software, the Haskell community has embraced licenses like (L)GPL, MIT and BSD, which are free software licenses. Actually the all-rights-reserved tag in Hackage [1] has only two packages tagged by it - the dummy no-op project HNop, and another package whose COPYING file contains a broken link and whose README says "BSD style license". Software freedom is an ethical basis for collaboration on making software that's truly good and loayl to its users, and providing them control and freedom to access and use their computing resources. It seems to me that the Haskell community is already enbracing this ethical basis, but Hackage doesn't provide any guarantees and it means that you'd have to check each package to be sure. By having that all-rights-reserved tag it also in a way welcomes software that doesn't go by these rules - however it seems that no packages do that even in the presence of the possibility. I'd like to make a suggestion: have Hackage accept only packages released under free software licenses. This is probably true for most/all packages there, but making it official will both send a strong message to and from the community, and provide people with the security and peace of mind, knowing their software is free as in freedom. It is also possible that companies use Haskell to create proprietary software using permissive-licensed libraries and tools from Hackage. I hope this isn't true, but even if it is, this software isn't offered by Hackage and I hope its existence doesn't affect the community's use of Hackage and free software. Would you consider to embrace free software officially, including in Hackage? Thanks for reading, waiting to hear from the community and from haskell.org maintainers, fr33domlover [1] http://hackage.haskell.org/packages/tag/all-rights-reserved --- fr33domlover <http://www.rel4tion.org/people/fr33domlover> GPG key ID: 63E5E57D (size: 4096) GPG key fingerprint: 6FEE C222 7323 EF85 A49D 5487 5252 C5C8 63E5 E57D

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Free monoids in Haskell
by Dr. Olaf Klinke 27 Feb '15

27 Feb '15

Dear Dan, you posted an entry on the Comonad.Reader blog about free monoids in Haskell. I commented about FM being a monad, and meanwhile verified that indeed every monoid is an FM-algebra. The proofs are analogous to proving that the continuation (_ -> r) -> r for fixed r is a monad and that r is an algebra for it, respectively. Moreover, the FM-algebra stucture is a monoid homomorphism w.r.t. the monoid instance you gave for (FM a). I'd like to ask what happens when one omits the Monoid constraint. That is, what are the elements of newtype Y a = Y { runY :: forall b. (a -> b) -> b } (Y a) is like (Control.ForFree.Yoneda Identity a), that is, elements of (Y a) should be natural transformations from the reader functor ((->) a) to the identity functor. If that is true, then the Yoneda lemma tells us that (Y a) is isomorphic to (Identity a), but at least operationally that seems not to be the case: u = runY (return undefined) has different semantics than u' = runY (Y undefined) as can be seen by applying both to const (). So return does not map ⊥ to (Y ⊥). I wonder whether one can exhibit other elements not equal to any return(x). Of course (Identity a) is actually the lifted a, since ⊥≠I dentiyt ⊥. Yet this does not serve as an explanation for u vs. u', as both u and u' are of the form (Y _), but one function evaluates its argument while the other does not. Olaf

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Kind synonyms
by Tom Ellis 27 Feb '15

27 Feb '15

Is there an extension that will allow me to do type T = (*, *) so I can replace type family F (k :: (*, *)) with type family F (k :: T) ? Tom

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Idiom Brackets for GHC (first full proposal)
by Oliver Charles 27 Feb '15

27 Feb '15

Hi all, A few days ago I made a post here to gauge interest in adding idiom brackets to GHC. Response was a bit more mixed than I was hoping, but no one seemed to drastically against the idea, so I've moved forward with a more detailed proposal. You can find the full proposal here: https://ocharles.org.uk/IdiomBrackets.html A particular difference in my proposal from existing solutions comes from my desire to lift almost *all* expressions - with the original syntax - into idiom brackets. This means normal function application and tuples, but also case expressions, let bindings, record construction, record update, infix notation, and so on. At first I was skeptical about this, but I am finding uses for this more and more. I really like how it lets me use the interesting data (that is, whatever is "under" the applicative functor) where it's most relevant - rather than having to build a function and thread that value back through. Examples of this can be seen in my proposal. To prove its use, I've been working with this Template Haskell expression: https://ocharles.org.uk/IdiomExp.hs It *almost* does exactly what I want, the only problem is I can't get Template Haskell to give me a difference between $(i [| const True False |]) and $(i [| (const True) False |]) which I was planning to be significant. The proposal has mostly grown into my own personal notebook, so I'm happy to clarify anything that is vague/contradictory/confusing. Look forward to hearing your thoughts! -- Ollie

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RankNTypes question
by Nikolay Amiantov 26 Feb '15

26 Feb '15

Hi Cafe, I can't find how can I do this: {-# LANGUAGE RankNTypes #-} example :: forall a. (forall b. b) -> a example = undefined The first `forall` is for cosmetic reasons. If I understand correctly this is a safe and legit function (please correct me if I'm wrong!). I thought that I can just define `example = id` and GHC would do the rest, but alas, that didn't happen. Thanks! -- Nikolay.

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Re: [Haskell-cafe] RankNTypes question
by Nikolay Amiantov 26 Feb '15

26 Feb '15

Ah, thanks! For some reason I haven't even thought of it -- I knew that `id` was wrong because of the type, but had no idea what else to try. Feeling stupid now. On 02/26/2015 09:46 PM, Marcin Mrotek wrote: > Well still, id is (a -> a). So given some function (forall b. b -> b) > it's result is a function (forall b. b -> b). This works: > > test :: (forall a. a -> a) -> Int -> Int > test f a = f a > > Kind regards, > Marcin Mrotek -- Nikolay.

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