Here's my take:
We want to define a function my-comp. It takes n functions and return
their composition.
We want to return a function of any number of arguments, so let's
start by working with a given
set of argument args, and returning the value of the composition
applied to those arguments.
- If I compose only one function, you get the value of the function itself:
(value-my-comp f) = (apply f args)
- By definition of composition:
(value-my-comp f1 f2 .... fn) = (f1 (value-my-comp f2 .... fn))
(Take all the arguments, apply it to the composition of function f2
.... fn, and then apply the result to f1).
Then you remove the dots:
(value-my-comp f1 & f2-fn) = (f1 (apply value-my-comp f2-fn))
Then you can just put them all together:
(fn [& fs] ;; we take a bunch of functions in
(fn [ & args ] ;; we take the arguments of the resulting function in.
(letfn [(value-my-comp ;; the value of the result in term of the
functions is:
([f] (apply f args)) ;; only one function
([f1 & f2-fn] (f1 (apply value-my-comp f2-fn))))] ;;
more than one function
(apply value-my-comp fs)))) ;; we compute the value of the
actual function we got.
If you don't like (apply value-my-comp f2-fn), you can make
value-my-comp actually takes a sequence
of functions:
(fn [& fs] ;; take the functions
(fn [ & args ] ;; the arguments of the resulting functions
(letfn [(value-my-comp
[[f1 & f2-fn]] ;; a list of functions
(if (empty? f2-fn) ;; only one function?
(apply f1 args) ;; then we apply the args to it
(f1 (value-my-comp f2-fn))))] ;; else we compose
(value-my-comp fs))))
This one is another way of writing the one you posted.
(Using a let instead of applying it directly, and not using a
non-varying argument a.)
If you want to see how this function can be written to be very
performant look at
comp source code, it is very interesting.
Best,
Nicolas.
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