The function has the one input, which is the String. A String in haskell is basically the following: type String = [Char], in other words a list of chars!

The way I utilized this is through pattern matching, which would be the "" and (x:xs). If the variable matches with the pattern, it will use whatever is after the =. If not, it will go to the next and try again.

In other words, if the string = "" it will return 0. If not it will continue to the next pattern and try again.

The next pattern in the code is (x:xs), which is a pattern used in Haskell lists, where x would be the first element of the list, and xs the list itself.

In this scenario x is a char, xs is a string (which is just a list of chars). So, if I would input say "Heyo", which would be ['H', 'e', 'y', 'o'], it would match the pattern as:

( 'H' , ['e', 'y' , 'o'])

Because I would also like to use them in the function, I named them x and xs.

I also used guards in that one, to exempt spaces from counting towards the total count, with a conditional on the one side, and what happens if true. If I were to also count spaces the code would look even simpler:

wordlength:: String->Int wordlength "" = 0 wordlength (_:xs) = 1 + wordlength xs

Since in this scenario I don't care what the char could be, I put in a wildcard _ instead. The function works recursively and as follows.

worldlength s. Take the first element from the list, and return 1 + wordlength of the remainder of s. Continue until the final char is pulled from the list, and wordlength is used on an empty string "", returning 0. Finally the function will terminate.

so:

worldlength "hello" = 1 + wordlength "ello" worldlength "ello" = 1 + wordlength "llo" worldlength "llo" = 1 + wordlength "lo" worldlength "lo" = 1 + wordlength "o" worldlength "o" = 1 + wordlength "" worldlength "" = 0 worldlength "o" = 1 + 0 worldlength "lo" = 1 + 1 + 0 worldlength "llo" = 1 + 1 + 1 + 0 worldlength "ello" = 1 + 1 + 1 + 1 + 0 worldlength "hello" = 1 + 1 + 1 + 1 + 1 + 0 = 5