No problem . Im strugelling to make acc work. I try to say that if the input list has only 1 item the outcome is the head of that list. But doing acc = Just (head a) or doing acc = Just (head acc) gives both that acc or a is not in scope also doing acc x = Just (head x) gives a error messages that the types are not matching. Roelof Stefan Höck schreef op 10-11-2014 15:23:
Just a quick note: That's 'typed holes' not 'type holes' ...
On Mon, Nov 10, 2014 at 03:12:20PM +0100, Stefan Höck wrote:
Hi Roelof
It seems like you try to do too many things at once. Here's how you could go about this step-by-step and let GHC help you implement your functions along the way:
First, give the type signature of your function:
last5 :: [a] -> Maybe a last5 = undefined
Now, load this into GHCi or compile with GHC. If it compiles, you're on the right track. Now, you want to implement it using a fold (try both, foldl and foldr):
last5 :: [a] -> Maybe a last5 xs = foldr _ _ xs
The underscores are 'type holes'. This tells the compiler to give you some information about what is supposed to be placed at the two positions. For the moment, we are only interested in the types of the things that go there. The compiler will tell you, that the hole at the first position is of type
a -> Maybe a -> Maybe a
and the hole at the second position is of type
Maybe a
Now, instead of filling the holes in place, let's define two helper functions together with their type signatures. You can later on inline them in your definition of last5, but for the time being, let's get as much help from the compiler as we can.
last5 :: [a] -> Maybe a last5 xs = foldr acc initial xs
acc :: a -> Maybe a -> Maybe a acc = undefined
initial :: Maybe a initial = undefined
Again, compile or load into GHCi. If you did anything wrong, the compiler will tell you so. There is only one possible way to implement function `initial` without cheating (= raising an error)
initial :: Maybe a initial = Nothing
Function `acc` can be implemented in several ways. Only one of them will lead to the desired behavior. Finding out the proper implementation is the main point of this folding-exercise. Try also an implementation using foldl. Does it behave as expected? What are the differences compared to foldr? When you feed your implementations a huge list - say [1..20000000] - what happens?
Note that whenever you get an error message in a rather complex function implementation, move local function definitions and lambdas to the top level, give them a type signature and implement them separately one at a time. Use type holes to let the compiler give assistance with type signatures and possible implementations. Once everything compiles and runs as expected, move the toplevel definitions back to where you'd like them best.
Stefan
PS: A more succint implementation of last5 would use currying:
last5 = foldr acc initial
PPS: If you get a stack overflow with very large lists, try using foldl' from Data.List (or better, once you learned about the Foldable type class, from Data.Foldable).
On Mon, Nov 10, 2014 at 02:28:08PM +0100, Roelof Wobben wrote:
I tried and so far I have this :
last5 = foldl(\acc x -> if x=[] then x else acc xs) [] xs
But now I get parse error on the -> part.
Roelof
Roelof Wobben schreef op 10-11-2014 13:47:
Thanks all,
I will try to make a fold solution as soon as I see that functional explained in the learnyouahaskell.
Roelof
Frerich Raabe schreef op 10-11-2014 11:43:
On 2014-11-10 11:16, Karl Voelker wrote:
On Mon, Nov 10, 2014, at 01:50 AM, Roelof Wobben wrote: > What do you experts think of the different ways ? 2 and 4 are quite similar, and both fine. 3 is not so good: guards provide weaker guarantees than patterns, and head and tail are partial functions. In addition to what Karl wrote, I'd like to suggest not using the semicolon all the time -- it's not needed and just adds noise.
All three implementations have in common that they do their own recursion. It would be a good exercise to try implementing last as a fold - in other words, letting the standard library do the recursion for you. Right - I suggest trying to express the problem with the most abstract function first, then consider using a fold, then use manual recursion. in your particular case you could exploit that for non-empty lists, getting the last element of a list is the same as getting the first element of a reversed list (and there are ready-made functions for reversing a list and getting the first element of a list).
_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners
_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners
Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners