# Untypeable Haskell 98

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(Difference between revisions)

NeilMitchell (Talk | contribs) |
DonStewart (Talk | contribs) (pattern annotatoins help) |
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− | + | Here we document code that looks like it should be valid Haskell98, but | |

− | + | isn't typeable without extensions: | |

+ | <haskell> | ||

f :: a -> a | f :: a -> a | ||

f x = g x | f x = g x | ||

Line 15: | Line 16: | ||

In GHC making the type of f :: forall a . a -> a, and adding -fglasgow-exts will make this code work. No such luck in Hugs. | In GHC making the type of f :: forall a . a -> a, and adding -fglasgow-exts will make this code work. No such luck in Hugs. | ||

+ | |||

+ | With pattern type annotations, however, the code works in both Hugs and GHC: | ||

+ | |||

+ | <haskell> | ||

+ | f :: a -> a | ||

+ | f (x :: a) = g x | ||

+ | where | ||

+ | g :: a -> a | ||

+ | g y = bind x y | ||

+ | |||

+ | bind :: a -> a -> a | ||

+ | bind a b = a | ||

+ | </haskell> |

## Revision as of 02:45, 29 August 2006

Here we document code that looks like it should be valid Haskell98, but isn't typeable without extensions:

f :: a -> a f x = g x where -- g :: a -> a g y = bind x y bind :: a -> a -> a bind a b = a

The above Haskell code is Haskell 98, rank-1 types, but cannot be given a type signature. Try commenting out the type signature for g and everything will go wrong.

In GHC making the type of f :: forall a . a -> a, and adding -fglasgow-exts will make this code work. No such luck in Hugs.

With pattern type annotations, however, the code works in both Hugs and GHC:

f :: a -> a f (x :: a) = g x where g :: a -> a g y = bind x y bind :: a -> a -> a bind a b = a