[plt-scheme] r6rs vs module language performance

From: Matteo Pradella (matteo.pradella at gmail.com)
Date: Fri Jun 27 15:05:33 EDT 2008

On Fri, Jun 27, 2008 at 6:01 PM, Matthew Flatt <mflatt at cs.utah.edu> wrote:
> At Fri, 27 Jun 2008 10:36:05 +0200, "Matteo Pradella" wrote:
>> On Fri, Jun 27, 2008 at 3:44 AM, Matthew Flatt <mflatt at cs.utah.edu> wrote:
>> > At Thu, 26 Jun 2008 06:05:23 -0700 (PDT), Matteo Pradella wrote:
>> >> I ported to r6rs a small research prototype I wrote a couple of years
>> >> ago (in mzscheme v.3, I think), to "try and feel" the new standard.
>> >> Just to waste some time, I tried on the same input file both the "pure
>> >> mzscheme" and the r6rs versions: the former took 5.8s on my machine,
>> >> while the latter 1m14s. The application core resides in a single r6rs
>> >> library, installed (and compiled) through plt-r6rs --install.
> It turns out to be an algorithmic difference.
> The `scheme/base' version uses `(foldl append ...)', and the R6RS
> version uses `(fold-left append ...)'. Those look the same, but it
> turns out that `foldl' and `fold-left' pass arguments to the given
> procedure in opposite orders. For the R6RS version, you want `(lambda
> (x y) (append y x))' instead of just `append'.
> Matthew

r6rs is surely full of surprises - thanks a lot for the tip!


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