Could someone help me make sense of this Num constraint given by GHCi? I was expecting to get: inc . const :: Num a => a -> b -> a rather than: inc . const :: Num (b -> a) => a -> b -> a Here's the transcript: GHCi, version 8.6.3: http://www.haskell.org/ghc/ :? for help Prelude> inc x = x + 1 Prelude> :t inc inc :: Num a => a -> a Prelude> :t const const :: a -> b -> a Prelude> :t inc . const inc . const :: Num (b -> a) => a -> b -> a Prelude> Thanks, Dimitri -- 2E45 D376 A744 C671 5100 A261 210B 8461 0FB0 CA1F
Remember that a -> b -> a is equivalent to a -> (b -> a) So: (forall a. Num a => a -> a) . (a -> (b -> a)) Becomes: (Num (b -> a). (b -> a) -> (b -> a)) . (a -> (b -> a)) Then: Num (b -> a). a -> (b -> a) Finally: Num (b -> a). a -> b -> a On 16/01/2019 15:52, Dimitri DeFigueiredo wrote:
Could someone help me make sense of this Num constraint given by GHCi? I was expecting to get:
inc . const :: Num a => a -> b -> a
rather than:
inc . const :: Num (b -> a) => a -> b -> a
Here's the transcript:
GHCi, version 8.6.3: http://www.haskell.org/ghc/ :? for help Prelude> inc x = x + 1 Prelude> :t inc inc :: Num a => a -> a Prelude> :t const const :: a -> b -> a Prelude> :t inc . const inc . const :: Num (b -> a) => a -> b -> a Prelude>
Thanks,
Dimitri
On 16/01/2019 15:52, Dimitri DeFigueiredo wrote:
...I was expecting to get: inc . const :: Num a => a -> b -> a
Prelude> inc x = x + 1
On Wed, Jan 16, 2019 at 5:06 PM Sylvain Henry
Remember that a -> b -> a is equivalent to a -> (b -> a)...
To augment Sylvain's explanation: The upshot of this unexpected type - which would almost certainly lead to a type error in any real program - is that you probably want const . inc rather than inc . const -Yitz
participants (3)
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Dimitri DeFigueiredo -
Sylvain Henry -
Yitzchak Gale