Question

If I use map or filter inside another function, does that function become a higher-order function? For example:

removeSpaces :: String -> String 
removeSpaces xs = filter (not . isSpace) xs 

Is removeSpaces a higher-order function?

Answer 1

No, removeSpaces is not a higher-order function.

A higher-order function is one that takes a function as an argument. removeSpaces doesn’t, so it isn’t.

Higher-orderness is a property of a function’s interface, not its implementation, so we can tell that removeSpaces isn’t higher-order just by looking at its type: it doesn’t take any functions as arguments, so it’s not higher-order. It is implemented with the use of a higher-order function, but that’s another matter entirely.

For example, filter is a higher-order function, because it is declared as taking a function as a parameter:

filter :: (a -> Bool) -> [a] -> [a]

But concat isn’t, because there aren’t any function types (a -> b) as arguments:

concat :: [[a]] -> [a]

Polymorphism poses a tricky edge-case: you can pass a function to id — id concat [[1, 2], [3, 4]] is [1, 2, 3, 4] — but its type does not declare it as taking any functions as arguments:

id :: a -> a

In this case, id is not higher-order. A function must explicitly have a function argument in its type to be higher-order.