On Mon, Dec 09, 2024 at 01:54:47PM +0000, Stefan Schmiedl wrote: > ------ Original Message ------ > > From to...@tuxteam.de > To guile-user@gnu.org > Date 09.12.2024 12:42:22 > Subject Re: sorted? > > > On Mon, Dec 09, 2024 at 11:37:33AM +0000, Ricardo G. Herdt wrote: > > > Hi Jeremy, > > > > > > Am 09.12.2024 11:21 schrieb Jeremy Korwin-Zmijowski: > > > > The reference says : > > > > > > > > Scheme Procedure: *sorted?* items less > > > > C Function: *scm_sorted_p* (items, less) > > > > > > > > Return |#t| if items is a list or vector such that, for each > > > > element x and the next element y of items, |(less y x)| returns > > > > |#f|. Otherwise return |#f|. > > > > > > > > I think the description should be : > > > > > > > > Return |#t| if items is a list or vector such that, for each element > > > > x and the next element y of items, |(less y x)| returns |#t|. > > > > Otherwise return |#f|. > > > > > > Actually no, since less is applied to y and x in that order. This way > > > (sorted? '(1 1) <) correctly returns #t as your experiments show. > > > > I don't get it. (< 1 1) is /always/ #f, regardless of the order of the > > ones? > > > > I'm as confused as Jeremy is. > > > > I understand the reference text as "Return #t if the list is _not > unsorted_". > Since (< 1 1) returns #f, '(1 1) is _not unsorted_ and all is well.
This seems the intention. But since it accepts an arbitrary "less" function, it ends being iffy. How do you go from some "less" to a "less-or-equal" without running into undecidability dark alleys? Cheers -- t
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