How do I remove every occurance of a value from a list in haskell using Prelude?

Question

I want to remove every occurance of a certain value from a list. I have written a function to do this:

removeall val [] = []
removeall val list = if (head list) == val
                     then removeall val (tail list)
                     else (head list):(removeall val (tail list))

but I would like to use Prelude if possible for reasons of elegance and readability.

Answer 1

removeall val list = filter (/= val) list

Answer 2

The following works as well

removeall val list = [ x | x <- list, x /= val ]

Answer 3

removeall = filter . (/=)

Answer 4

This is just a rewrite of yours which removes the head and tail function calls.

removeall val [] = []
removeall val (x:xs) = if (x == val) 
                         then removeall val xs 
                         else x:removeall val xs

Personally I prefer the

removeall = filter . (/=)

one given by the others but that might be harder for a beginner to understand quickly.