Re: [Haskell-cafe] defining last using foldr

You can consider foldr to be continual modification of a state. The initial state is given as an argument, and then the (a -> b -> b) function is passed the next element of the list and the current state, and it returns the new state. foldr will then return the final state, from which the result can be extracted. Does that help? On 14/08/07, Alexteslin wrote:

Well, i have tried cons (:) operator but when it passed to foldr doesn't work because cons operator operates first character and then the list but the foldr argument takes a function (a->a->a). Maybe i am missing the point here?

Aaron Denney wrote:

...

On 2007-08-14, Alexteslin wrote:

...

Hi,

I am trying to do the exercise which asks to define built-in functions 'last' and 'init' using 'foldr' function, such as last "Greggery Peccary" = 'y'

the type for my function is:

myLast :: [Char] -> Char

I am not generalizing type so that make it less complicated. But what ever i am trying would not work. The only function type foldr takes as an argument is either (a->a->a) or (a->b->b) and none of the functions i found that would match this type from Char to Char. So in other words should be (Char->Char-Char). I can define the function without foldr but that misses the point of the exercise.

Folds replace the "cons" operator (:) with the function you pass it. If you want the tail of the list, you want what is on the right hand side of every cons (unless that's []).

-- Aaron Denney -><-

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