[Haskell-cafe] Deducing a type signature
Bird 1.6.3 requires deducing type signatures for the functions strange and stranger. Are my solutions below correct? (i) strange f g = g (f g) Assume g :: a - b. Then f :: (a - b) - c. But since g :: a - b,f g :: a, so c = a. Therefore, f :: (a - b) - a, and g (f g) :: a.Therefore, strange :: ((a - b) - a) - (a - b) - a. (ii) stranger f = f f Assume f :: a - b. Since f f is well-typed, type unification requiresa = b. Therefore, f :: a - a, and stranger :: (a - a) - a. _ Hotmail is redefining busy with tools for the New Busy. Get more from your inbox. http://www.windowslive.com/campaign/thenewbusy?ocid=PID28326::T:WLMTAGL:ON:WL:en-US:WM_HMP:042010_2___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Deducing a type signature
On May 20, 2010, at 11:03 AM, R J wrote: stranger f = f f This doesn't have a type in Haskell. Suppose f :: a - b Then if f f made sense, a = (a - b) would be true, and we'd have an infinite type. Type the definition into a file, and try loading it into ghci: Occurs check: cannot construct the infinite type: t = t - t1 Probable cause: `f' is applied to too many arguments In the expression: f f In the definition of `stranger': stranger f = f f ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Deducing a type signature
(i) strange f g = g (f g) Assume g :: a - b. Then f :: (a - b) - c. But since g :: a - b, f g :: a, so c = a. Therefore, f :: (a - b) - a, and g (f g) :: a. Therefore, strange :: ((a - b) - a) - (a - b) - a. Almost. The return type of strange is the same as the return type of g (the outermost function), namely b. So strange :: ((a - b) - a) - (a - b) - b. Dan R J wrote: Bird 1.6.3 requires deducing type signatures for the functions strange and stranger. Are my solutions below correct? (i) strange f g = g (f g) Assume g :: a - b. Then f :: (a - b) - c. But since g :: a - b, f g :: a, so c = a. Therefore, f :: (a - b) - a, and g (f g) :: a. Therefore, strange :: ((a - b) - a) - (a - b) - a. (ii) stranger f = f f Assume f :: a - b. Since f f is well-typed, type unification requires a = b. Therefore, f :: a - a, and stranger :: (a - a) - a. Hotmail is redefining busy with tools for the New Busy. Get more from your inbox. See how. http://www.windowslive.com/campaign/thenewbusy?ocid=PID28326::T:WLMTAGL:ON:WL:en-US:WM_HMP:042010_2 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
