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Today's Topics:
1. Re: My Continuation doesn't typecheck (David McBride)
2. Re: My Continuation doesn't typecheck (martin)
3. Bool newtype (Imants Cekusins)
----------------------------------------------------------------------
Message: 1
Date: Sat, 6 Aug 2016 13:01:00 -0400
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] My Continuation doesn't typecheck
Message-ID:
<CAN+Tr43tech2-29oECPAWjMEkdX2cA+m=mVHdGH+v6_-=z4...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
The only way to do this is to do it step by step.
:t combine
combine :: t1 -> (t1 -> t2 -> t) -> t2 -> t
>:t combine 9
combine 9 :: Num t1 => (t1 -> t2 -> t) -> t2 -> t
>:t f1
f1 :: Int -> (Integer -> r) -> r
>:t combine 9 f1
combine 9 f1 :: (Integer -> t) -> t
>:t f2
f2 :: Integer -> (String -> r) -> r
>:t combine 9 f1 f2
combine 9 f1 f2 :: (String -> r) -> r
At some point the t2 in combine becomes a function, which causes the rest
of the type to change. I feel like combine was meant to be something else,
f (g a) or g (f a) or something else, but I'm not sure what.
On Sat, Aug 6, 2016 at 4:03 AM, martin <[email protected]> wrote:
> Hello all,
>
> in order to gain some intuition about continuations, I tried the following:
>
> -- two functions accepting a continuation
>
> f1 :: Int -> (Integer->r) -> r
> f1 a c = c $ fromIntegral (a+1)
>
> f2 :: Integer -> (String -> r) -> r
> f2 b c = c $ show b
>
> -- combine the two functions into a single one
>
> run1 :: Int -> (String -> r) -> r
> run1 a = f1 a f2
>
>
> -- *Main> run1 9 id
> -- "10"
>
> So far so good.
>
>
> Then I tried to write a general combinator, which does not have f1 and f2
> hardcoded:
>
> combine a f g = f a g
>
> -- This also works
>
> -- *Main> combine 9 f1 f2 id
> -- "10"
>
>
> What confuses me is the the type of combine. I thought it should be
>
> combine :: Int ->
> (Int -> (Integer->r) -> r) -> -- f1
> (Integer -> (String -> r) -> r) -> -- f2
> ((String -> r) -> r)
>
>
> but that doesn't typecheck:
>
> Couldn't match expected type ‘(String -> r) -> r’
> with actual type ‘r’
>
>
> Can you tell me where I am making a mistake?
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 2
Date: Sun, 7 Aug 2016 13:05:02 +0200
From: martin <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] My Continuation doesn't typecheck
Message-ID: <[email protected]>
Content-Type: text/plain; charset=utf-8
David,
I used your method of hardcoding some of the parameters to find the correct
type of 'combine'. It is not at all what I
expected or wanted, but here it is:
combine :: Int -> (Int -> (Integer -> (String -> r) -> r) -> (String -> r) -> r
-> String) ->
(Integer -> (String -> r) -> r) -> (String -> r) -> r
-> String
Not sure what this is trying to tell me.
Am 08/06/2016 um 07:01 PM schrieb David McBride:
> The only way to do this is to do it step by step.
> :t combine
> combine :: t1 -> (t1 -> t2 -> t) -> t2 -> t
>
>>:t combine 9
> combine 9 :: Num t1 => (t1 -> t2 -> t) -> t2 -> t
>
>>:t f1
> f1 :: Int -> (Integer -> r) -> r
>>:t combine 9 f1
> combine 9 f1 :: (Integer -> t) -> t
>
>>:t f2
> f2 :: Integer -> (String -> r) -> r
>>:t combine 9 f1 f2
> combine 9 f1 f2 :: (String -> r) -> r
>
> At some point the t2 in combine becomes a function, which causes the rest of
> the type to change. I feel like combine
> was meant to be something else, f (g a) or g (f a) or something else, but I'm
> not sure what.
>
>
> On Sat, Aug 6, 2016 at 4:03 AM, martin <[email protected]
> <mailto:[email protected]>> wrote:
>
> Hello all,
>
> in order to gain some intuition about continuations, I tried the
> following:
>
> -- two functions accepting a continuation
>
> f1 :: Int -> (Integer->r) -> r
> f1 a c = c $ fromIntegral (a+1)
>
> f2 :: Integer -> (String -> r) -> r
> f2 b c = c $ show b
>
> -- combine the two functions into a single one
>
> run1 :: Int -> (String -> r) -> r
> run1 a = f1 a f2
>
>
> -- *Main> run1 9 id
> -- "10"
>
> So far so good.
>
>
> Then I tried to write a general combinator, which does not have f1 and f2
> hardcoded:
>
> combine a f g = f a g
>
> -- This also works
>
> -- *Main> combine 9 f1 f2 id
> -- "10"
>
>
> What confuses me is the the type of combine. I thought it should be
>
> combine :: Int ->
> (Int -> (Integer->r) -> r) -> -- f1
> (Integer -> (String -> r) -> r) -> -- f2
> ((String -> r) -> r)
>
>
> but that doesn't typecheck:
>
> Couldn't match expected type ‘(String -> r) -> r’
> with actual type ‘r’
>
>
> Can you tell me where I am making a mistake?
>
> _______________________________________________
> Beginners mailing list
> [email protected] <mailto:[email protected]>
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> <http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners>
>
>
>
>
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
------------------------------
Message: 3
Date: Sun, 7 Aug 2016 13:45:54 +0200
From: Imants Cekusins <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: [Haskell-beginners] Bool newtype
Message-ID:
<CAP1qinZfZDp-nU_iN-5BZo1zw6a=ekeboegc8_-aevl3cgt...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
for a Bool-like newtype:
newtype B = B Bool
, is there an easy way to use this newtype B in place of Bool?
e.g.
let b1 = B True
in if b1 then 1 else 0
?
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