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This is pretty inefficient haskell, as you have to traverse the entire list and make a copy of it each time you use the ++ operator. A more efficient solution is rev = foldl (\acc x -> x : acc) []. this way, you avoid the expense of using the ++ operator
Although, this is pretty advanced since we're making use of higher order functions already.
Deriving the 'inefficient' solution first, before learning why it's inefficient is the way to go.
yet I read on haskell.org/prelude, "Also note that if you want an efficient left-fold, you probably want to use foldl' instead of foldl." So maybe this isn't that efficient? Can you say a word about this? (I'm a complete newbie... :-))
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This is pretty inefficient haskell, as you have to traverse the entire list and make a copy of it each time you use the ++ operator. A more efficient solution is
rev = foldl (\acc x -> x : acc) []. this way, you avoid the expense of using the ++ operatorOr, with
flip:Although, this is pretty advanced since we're making use of higher order functions already. Deriving the 'inefficient' solution first, before learning why it's inefficient is the way to go.
yet I read on haskell.org/prelude, "Also note that if you want an efficient left-fold, you probably want to use
foldl'instead offoldl." So maybe this isn't that efficient? Can you say a word about this? (I'm a complete newbie... :-))