May 27, 2020 11:33:58 PM CEST Tony Morris wrote:foldl does a loop, foldr does constructor replacement https://www.youtube.com/watch?v=GPwtT31zKRY On 5/28/20 4:55 AM, Alexander Chen wrote:

Hi,

I guess it's time to get acquainted with foldr and foldl.

*prelude>xs = [1..5] ++ undefined* *prelude> foldr const 0 xs* 1

I assume it goes something like this:

( 1 `const`(2 `const`(3 `const`(4 `const`(5 `const`(undefined `const` 0))))))                                                               ( 5 `const` undefined)                                                (4 `const` 5)                                     (3 `const` 4)                 ( 2 `const` 3) (1 `const`2) = 1

========================================================================

What i don't get is the opposite:

*prelude> foldl const 0 xs* error

in my mind this should go like this:  ((((((0`const`1)`const` 2) `const` 3 )`const` 4)`const`  5) `const` undefined)                     (0 `const`2)                                       (0`const` 3)                                                                   ( 0`const`4)                                                                        (0`const` 5)                                                                                       (0 `const` undefined ) = 0

I have been told that the main difference between foldl and foldr is that foldl needs to evaluate the whole spline before it continues. And i guess that has something to do with it. What I don't understand is WHY foldl need to do this and foldr doesn't.

thanks in advance!

best,

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners