Working my way through the Clojure Koans I learned about Higher Order Functions.
The koan meditation was this:
"Higher-order functions take function arguments" (= 25 (___ (fn [n] (* n n))))
The question is: what’s an higher-order function?
Wikipedia tells me that an HOF is a function that does at least one of these:
- takes one (or more) functions as an input
- returns another function as an output
An example from calculus is the derivative function.
Also all other functions are called first-order functions.
The solution to the koan above is know clearer: it needs to be a function that takes another function as argument, in this case a square functions, and applies to a number in order to output 25.
(def function_of_five (fn [f] (f 5))
So when we write:
(= 25 ((fn [f] (f 5)) (fn [n] (* n n)))))
What happens it that the
(fn [n] (* n n)) and the
5, which results into
A deeper look at Clojure’s HOF can be found here.
I’m getting more and more in love with Clojure. I like how it’s so different from what I’m used to as an OO developer. And I like how it’s slowly bringing back some of the academic mathematic concepts I got in touch with (learned would be an overstatement) at uni and that I considered useless for my day to day job for a long time.