Implement the

replaceHead :: [a] -> [a] -> [a]

function, that works as the title says. For example:

replaceHead "okay" "mom" == "mkay"

or:

replaceHead [0,2,3] [1,20,30] == [1,2,3]

I tried using

replaceHead a b = [head b, a]

which doesn't compile. I don't have any more ideas as to how I can take b's first element, swap it with a's first element, and then return that list. Can you help? Thanks for you answers in advance!

CodePudding user response：

The problem is that a is a list of items, whereas head b is an item. You can thus not use [head b, a] since then the first item is for example an Int, whereas a is an [Int]. In order to meet the sample output, we furthermore need to work with the tail of the first list, not the entire first list.

You can however prepend the tail a list by using the (:) :: a -> [a] -> [a] data constructor:

replaceHead :: [a] -> [a] -> [a]
replaceHead a b = head b : tail a

It might be better to work with pattern matching here, and implement this as:

replaceHead :: [a] -> [a] -> [a]
replaceHead (_:as) (b:_) = b : as

In that case you probably should define a case when the first and/or second list is empty:

replaceHead :: [a] -> [a] -> [a]
replaceHead (_:as) (b:_) = b : as
replaceHead [] [] = …
replaceHead [] (b:bs) = …
replaceHead (a:as) [] = …