Re: [Haskell-cafe] const is question

On Jun 14, 2008, at 11:14 AM, odtaa48 wrote:

const id 0 5 -> 5 -- ie. 'takes' the 'last' one

The type of const is a -> b -> a. The first argument to const here is id and the second is 0, so the evaluation of const id 0 is going to result in id. Then we apply id to 5, yielding 5. Perhaps it would be more clear to write the expression is (const id 0) 5.

const (id 0) 5 -> 0 -- ie. 'takes' the 'first' one

Here the two arguments to const are (id 0) and 5. Since the effect of const is to toss out the second argument, this evaluates to id 0, which evaluates to simply 0. Does that help? - Jake McArthur