[Haskell-beginners] A question about an infinite type

[Lorenzo Bolla]

You can find your answer here:
http://en.wikibooks.org/wiki/Haskell/YAHT/Type_advanced



On Tue, Feb 28, 2012 at 12:52 PM, Costello, Roger L. <costello at mitre.org>wrote:

> Hi Folks,
>
> Here is an interesting phenomena:
>
> let f = (\x y -> x y x)
>
> let p = (\x y -> 1)
>
> Now let's evaluate  f p 1
>
> f p 1  = (\x y -> x y x) p 1                -- by replacing f with its
> definition
>
>          = p 1 p                                        -- by substituting
> x with p and y with 1
>
>          = (\x y -> 1) 1 p                     -- by replacing the first p
> with its definition
>
>          = 1                                               -- the function
> returns 1 regardless of its arguments
>
> However, Haskell will not proceed with that evaluation because it will
> first determine the
>
> type signature of f and judge it to be an infinite type.
>
> Conversely, if f is omitted and this expression
>
>    p 1 p
>
> is evaluated then the result 1 is generated.
>
> Why does f p 1 (which evaluates to p 1 p) fail whereas p 1 p succeeds?
>
> What lesson should I learn from this example?
>
> /Roger
>
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