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Today's Topics:
1. unwanted Fractional constraint (M Douglas McIlroy)
2. Re: unwanted Fractional constraint (David McBride)
3. Re: unwanted Fractional constraint (David James)
----------------------------------------------------------------------
Message: 1
Date: Fri, 11 Dec 2020 08:03:03 -0500
From: M Douglas McIlroy <m.douglas.mcil...@dartmouth.edu>
To: beginners@haskell.org
Subject: [Haskell-beginners] unwanted Fractional constraint
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<cakh6pivhxddtx4exanfjk23fqm2geocoix-ymwg_kymf8nq...@mail.gmail.com>
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For rational functions that take on integer values at integer
arguments, for example n*(n+1)/2, is there a way to doctor the
corresponding Haskell definition
f n = n*(n+1)/2
so that the type signature becomes
f :: Num a => a -> a
rather than
f :: Fractional a => a -> a
Doug McIlroy
------------------------------
Message: 2
Date: Fri, 11 Dec 2020 08:42:03 -0500
From: David McBride <toa...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] unwanted Fractional constraint
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<CAN+Tr42FueMGDqwg+ueZ9emk-kUU7PC_MnqFA-kFB3OARYz=v...@mail.gmail.com>
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The problem is that the moment you divided by two, it can no longer be an
instance of Num.
Instances of Num include: Integers, Ints and Words. Once divided, you can
no longer use that result in places where an Integer is required, because
it won't be.
What you can do is use the `div` function which will round down to the
nearest Integer value. Then it is an instance of Integral, which includes
Integers, Ints, and Words (but floating point types)
You can also use `round`, `floor`, or `ceiling` to round your result to an
appropriate integer after you've divided.
On Fri, Dec 11, 2020 at 8:04 AM M Douglas McIlroy <
m.douglas.mcil...@dartmouth.edu> wrote:
> For rational functions that take on integer values at integer
> arguments, for example n*(n+1)/2, is there a way to doctor the
> corresponding Haskell definition
>
> f n = n*(n+1)/2
>
> so that the type signature becomes
>
> f :: Num a => a -> a
>
> rather than
>
> f :: Fractional a => a -> a
>
> Doug McIlroy
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Fri, 11 Dec 2020 16:38:46 +0000
From: David James <dj112...@outlook.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] unwanted Fractional constraint
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<am6pr04mb4214c1730b0fcf72c079bf16b6...@am6pr04mb4214.eurprd04.prod.outlook.com>
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[Attempting to resend]
Hi - I think your logic is:
I can define these two:
fFrac :: Fractional a => a -> a
fFrac n = n * (n+1) / 2
fInt :: Integral a => a -> a
fInt n = n * (n+1) `div` 2
so that, e.g.
fFrac (5.0 :: Float) == 15.0 :: Float
fInt (5 :: Integer) == 15 :: Integer
And all number types are either Integral or Fractional, so surely I should be
able to define a single function of type:
f :: Num a => a -> a
This would seem reasonable, but I think there’s a problem with the last
assumption. It is indeed possible for other types to be instances of Num, but
not of Integral or Fractional.
For example, I could define:
instance Num Bool where
fromInteger 0 = False
fromInteger _ = True
(+) = (&&)
(*) = (||)
abs = id
signum _ = True
negate = not
Now this would probably be pretty dumb (and probably doesn’t comply with
expectations<https://hackage.haskell.org/package/base-4.14.0.0/docs/Prelude.html#t:Num>),
but is possible. (And also pretty dumb to define it without also define an
instance Integral Bool where ..., but still possible).
So I don’t think
f :: Num a => a -> a
could be possible, since Num by itself (& the dumb Bool instance) has no way to
do the division. (At least that I can think of, but would be very interested to
hear if there is).
Regards, David.
From: Beginners <beginners-boun...@haskell.org> on behalf of M Douglas McIlroy
<m.douglas.mcil...@dartmouth.edu>
Sent: Friday, December 11, 2020 1:03:03 PM
To: beginners@haskell.org <beginners@haskell.org>
Subject: [Haskell-beginners] unwanted Fractional constraint
For rational functions that take on integer values at integer
arguments, for example n*(n+1)/2, is there a way to doctor the
corresponding Haskell definition
f n = n*(n+1)/2
so that the type signature becomes
f :: Num a => a -> a
rather than
f :: Fractional a => a -> a
Doug McIlroy
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