(Add description for each type of solution. Change formatting, ensure that all code is enclosed in "haskell" braces.)
Revision as of 13:04, 15 May 2014
(*) Find the number of elements of a list.
The simple, recursive solution.This is similar to the
myLength :: [a] -> Int myLength  = 0 myLength (_:xs) = 1 + myLength xs
The prelude for haskell 2010 can be found here.
Same, but now we use an "accumulator" argument.
myLength :: [a] -> Int myLength list = myLength_acc list 0 where myLength_acc  n = n myLength_acc (_:xs) n = myLength_acc xs (n + 1)
myLength :: [a] -> Int myLength1 = foldl (\n _ -> n + 1) 0 myLength2 = foldr (\_ n -> n + 1) 0 myLength3 = foldr (\_ -> (+1)) 0 myLength4 = foldr ((+) . (const 1)) 0 myLength5 = foldr (const (+1)) 0 myLength6 = foldl (const . (+1)) 0
We can also create an infinite list starting from 1. Then we "zip" the two lists together and take the last element (which is a pair) from the result:
myLength :: [a] -> Int myLength1 xs = snd $ last $ zip xs [1..] -- Just for fun myLength2 = snd . last . (flip zip [1..]) -- Because point-free is also fun myLength3 = fst . last . zip [1..] -- same, but easier
We can also change each element into our list into a '1' and then add them all together.
myLength :: [a] -> Int myLength = sum . map (\_->1)