99 questions/Solutions/4

(*) Find the number of elements of a list.

The simple, recursive solution. This is similar to the length from Prelude:

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)

Using foldl/foldr:

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)