Hi Michael, michael rice wrote:
as opposed to an "inferred type"?
Can you deduce from the following example?
Prelude> let foo = () :: Int
<interactive>:1:10:
Couldn't match expected type `Int' against inferred type `()'
In the expression: () :: Int
In the definition of `foo': foo = () :: Int
Hope this helps! Martijn. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
