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John.Cowan wrote:
For example, if the programmer knows that f is a list of fixnums, there is no way to apply fixnum addition to them, whereas (apply fx+ f) is trivial.
Well. fx+ as currently specified is binary.
So (apply fx+ f) is equivalent to:
(apply (lambda ((v1 :: <fixnum>) (v2 :: <fixnum>)) (+ v1 v2))
f)
For a general-length list one can of course use reduce.
Admittedly more tedious.
Similarly, a HOF involving fixnums can't take advantage of fixnum arithmetic: (define (op f (x :: fixnum) (y :: fixnum)) (f x y)) will perform generic addition, not fixnum addition, if called as (op + 2 3).
No, it will perform fixnum arithmetic, but fixnum arithmetic will be selected using a run-time type dispatch. This requires that integers and fixnums are distinguishable at runtime. I.e. (eqv? 10 (as <fixnum> 10)) => #f though: (= 10 (as <fixnum> 10)) => #t just like: (eqv? 10 10.0) =>f and (= 10 10.0) => #t. I have mixed feelings about whether fx+ should be in R6RS. I would like for (fx+ x y) to be equivalent to: (+ (as <fixnum> x) (as <fixnum) y)) That does require that 10 and (as <fixnum> 10) be different. This can be partly hidden from implementations that don't have separate fixnum values, if we define (fixnum? x) as testing whether x is in the fixnum *range* rather than having the fixnum type. What I do need to request is this: (eqv? 5 (fx+ 2 3)) => unspecified -- --Per Bothner per@xxxxxxxxxxx http://per.bothner.com/