How to iterate through a list, element by element

We can use Maybe to model whether the index was valid.

nth :: Int -> [a] -> Maybe a
nth 0 (x : _) = Just x
nth n (x : xs) = nth (n - 1) xs
nth _ [] = Nothing

We can pattern match on the index to get our base case, and the list to get the first element and tail.

What you're doing there with (_:x) is called "pattern matching" in case you didn't know. The general pattern for iterating through a list would be (x : xs) where x is head element of the list being matched and xs is the rest of the list. If you use _ you don't remove anything it is still matched to _ which is the convention for saying "I won't use this".

With that you can make a function like this:

myNth :: [a] -> Int -> a
myNth  []      _ = error "out of range"
myNth (x : xs) 0 = x
myNth (_ : xs) n = myNth xs (n - 1)

Whenever myNth is called it will go top to bottom over those definitions trying to match the patterns to the input. So when you call myNth [10,11] 1 it won't match the first clause because [10,11] doesn't match an empty list, it won't match the second either because 1 is not 0 and so it will match the third case where it will match the [10,11] on (10 : [11]), therefore _ is 10 and xs is [11] and 1 will be matched as n. Then it calls itself recursively, as myNth [11] 0. Now that will match the second case and it will return x from the match of [11] on (11 : [])

Like 414owen said you can use the Maybe a type to avoid using error.

P.S.: I don't know how beginner you are but I assume you know of the : operator, it prepends an element to a list... If you go more in depth (afaik) every list is actually stored as a sequence of a:(b:(c:(d:(e:[])))) which is equivalent to [a,b,c,d,e] which is equivalent to a:[b,c,d,e] etc.