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. 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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