Re: a type question

rui yang wrote:

Suppose I have a function:

funcmap :: a->b->c

can I use type synonyms to describe a new type like this: Type Funcmap = a->b>c ?

First, it's 'type' not 'Type'. Second, you want '->' not '>'. Third, all type variables in the RHS must be on the LHS. So, we get type Funcmap a b c = a->b->c