How to do something to all elements of a list but the last

In this post on the PLT discussion list I sleepily wondered how to do something to all elements of a list but the last.

The reasons I asked is because I had used ‘do’ loops in lieu of something better, and from what I can gather, you basically never write ‘do’ loops in Scheme (maybe once every 30 years you do use ‘do’).

I hastily posted a solution using the ‘match’ library as I had just read the documentation while cutting over from ‘mzlib/match’ to ‘match’. Both Robby and Ethan replied with much, much faster solutions. Have a look:

```
#lang scheme

(require (prefix-in srfi/1: srfi/1))

; Grant's
(define (foo1 args fun last-fun)
(match args
[(list arg args ..1)
(fun arg)
(foo1 args fun last-fun)]
[(list arg)
(last-fun arg)]))

(provide/contract
[foo2 (-> (cons/c any/c (listof any/c))
(-> any/c any)
(-> any/c void?)
void?)])

; Robby's
(define (foo2 args fun last-fun)
(let loop ([fst (car args)]
[rst (cdr args)])
(cond
[(empty? rst) (last-fun fst)]
[else (fun fst) (loop (car rst) (cdr rst))])))

(define data (srfi/1:make-list 10000 "hi"))

; Ethan's
(define (foo3 lst fun funl)
(begin
(for-each fun (srfi/1:drop-right lst 1))
(funl (srfi/1:take-right lst 1))))

(time
(foo1 data
(λ (x) (void))
(λ (x) (void))))

(time
(foo2 data
(λ (x) (void))
(λ (x) (void))))

(time
(foo3 data
(λ (x) (void))
(λ (x) (void))))

> cpu time: 2593 real time: 2625 gc time: 718
> cpu time: 0 real time: 0 gc time: 0
> cpu time: 0 real time: 0 gc time: 0

```

In the end, I only needed this functionality in two places, and I probably would not have required it if I had designed things better in the first place.