I solved the 58th 4clojure problem using recursion but then I looked at another persons solution and found this:
(fn [& fs] (reduce (fn [f g] #(f (apply g %&))) fs))
Which is more elegant than my solution. But I don't understand what %& means? (I do understand what % means but not when it's combined with &). Could anyone shed some light on this?
It means the "rest arguments", as per this source.
Arguments in the body are determined by the presence of argument literals taking the form %, %n or %&. % is a synonym for %1, %n designates the nth arg (1-based), and %& designates a rest arg.
Note that the & syntax is reminiscent of the & more arguments in function parameters (see here), but &% works inside an anonymous function shorthand.
Some code to compare anonymous functions and their anonymous function shorthand equivalent :
;; a fixed number of arguments (three in this case)
(#(println %1 %2 %3) 1 2 3)
((fn [a b c] (println a b c)) 1 2 3)
;; the result will be :
;;=>1 2 3
;;=>nil
;; a variable number of arguments (three or more in this case) :
((fn [a b c & more] (println a b c more)) 1 2 3 4 5)
(#(println %1 %2 %3 %&) 1 2 3 4 5)
;; the result will be :
;;=>1 2 3 (4 5)
;;=>nil
Note that the & more or %& syntax gives a list of the rest of the arguments.
