Scheme Macro for nesting expressions

Question

Can a macro be written in Scheme (with define-syntax, for example) which will take expressions like this:

(op a b c d e f g h i j)

And yield expressions like this as output?

(op (op (op (op (op (op (op (op (op a b) c) d) e) f) g) h) i) j)

Of course, for arbitrary lengths. I can't think of a way to do it, given some template like this:

(define-syntax op
  (syntax-rules ()
    [(_) 'base-case]
    [(v1 v2 ...) 'nested-case??]))

I am curious. Was this just a question of theoretical interest or do you have an interesting use case? In general, that kind of wrapping would be achieved by a foldLeft. — namin, Dec 04 '08 at 20:34
hmm, good point. i mostly asked this because i didn't think of your way =P. i didn't realize this is exactly what foldl does. — Claudiu, Dec 10 '08 at 20:32

namin · Accepted Answer · 2008-12-04T13:05:00.107

6

(define bop list)

(define-syntax op
  (syntax-rules ()
    ((op a b) (bop a b))
    ((op a b c ...) (op (bop a b) c ...))))

For example, (op 1 2 3 4) expands to (bop (bop (bop 1 2) 3) 4) and evaluates to (((1 2) 3) 4).

edited Dec 04 '08 at 13:05

answered Dec 04 '08 at 11:40

namin

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1

Why are you using bop in the literal identifiers if you are not matching it? – leppie Dec 04 '08 at 12:09
1

scheme macros are really really smart with their ... aren't they? – Claudiu Dec 04 '08 at 18:44

score 1 · Answer 2 · answered Dec 10 '08 at 20:27

The function you want to apply to the arguments should itself be an argument to the macro. Barring that, my solution was the same.

#!r6rs

(import (rnrs base))

(define-syntax claudiu
  (syntax-rules ()
    ((claudiu fun first second)
     (fun first second))
    ((claudiu fun first second rest ...)
     (claudiu fun (claudiu fun first second) rest ...))))

score 0 · Answer 3 · answered Dec 04 '08 at 18:52

To show how the answer works out:

(op 1 2 3 4)

This is an op with 4 statements, so the 2nd case gets selected with a=1, b=2, c=3, ...=4:

(op (bop 1 2) 3 4)

This is an op with 3 statements, so 2nd case again. a=(bop 1 2), b=3, c=4:

(op (bop (bop 1 2) 3) 4)

Now this is a bop with 2 statements, so a=(bop (bop 1 2) 3), b=4, and it's done.

Scheme Macro for nesting expressions

3 Answers3

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