Mark H Weaver <m...@netris.org> writes: > David Kastrup <d...@gnu.org> writes: > >> For all the cleverness involved here, one has to run through the whole >> list anyway. It makes no real sense to do this in this manner. The >> motivation may be to have a warm cache when k is small, but the result >> is self-defeating because of VM stack buildup. >> >> (define (drop-right lis k) >> (drop lis (- (length lis) k))) >> >> should be all that is needed. > > That won't be sufficient. SRFI-1 specifies that 'drop-right' works > for dotted lists, i.e. finite non-nil-terminated lists, whereas > 'length' accepts only proper lists.
Yes, I noticed. > It includes these examples: > > (drop-right '(1 2 3 . d) 2) => (1) > (drop-right '(1 2 3 . d) 0) => (1 2 3) > > See <http://srfi.schemers.org/srfi-1/srfi-1.html>. > > Would you like to propose another fix? The simplest fix would be using length+ rather than length, but that would require length+ to return the length of dotted lists (defined as its spine) rather than #f. As I interpret the standard, length+ for dotted lists is unspecified. It now returns #f. Other options would be throwing an error, delivering the length of the spine, or returning the negative of the total number of elements (meaning that the "dotted list" 5 has a length+ of -1, distinguishable from (length+ '())). Since it is suggested in srfi-1 that many routines do something useful when given dotted lists, it's rather inconvenient that there is _no_ list length operator working on dotted lists. There are routines that use length+ as a building block in order to admit circular lists and they tend to fail with surprising error messages when given dotted lists. So I lean towards making length+ put out the spine length of dotted lists, obviously requiring a review of its few uses. Having yet another length operator, in contrast, seems like overkill. While a "arithmetic if" style operator yanking out negative values for dotted lists would have the advantage of delivering complete information, its usage would, well, be awkward. And in explicit recursion/loops, one will eventually arrive at the end of the processed list and will see whether the first non-pair is '() or not without additional cost. -- David Kastrup
