[Haskell-beginners] Re: proper way to read fold types

[Brandon S Allbery KF8NH]

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On 7/26/10 07:30 , dan portin wrote:
> I'm still a Haskell newbie, so take this with a grain of salt: the type
> signature of /foldr/ and /foldr1/ confuses me also. While I understand how
> each operates, I've never found a description of the type signatures of
> either functions which explains why the function argument of /foldr1/ is of
> type (a -> a -> a) instead of (a -> b -> b). This is the conclusion I came

foldr1 is just foldr where "z" is taken off the end of the list.  Therefore,
given [a] as the list type, b (the type of z) unifies with a; and since the
folding function is initially called with z along with the first element of
the list, (a -> b -> b), in the case of foldr1 it is forced to (a -> a -> a).

- -- 
brandon s. allbery     [linux,solaris,freebsd,perl]      allbery at kf8nh.com
system administrator  [openafs,heimdal,too many hats]  allbery at ece.cmu.edu
electrical and computer engineering, carnegie mellon university      KF8NH
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