yea On Fri, Feb 10, 2017 at 5:21 PM, Joh-Tob Schäg <johtob...@gmail.com> wrote:
> (de natsum (N) > (if (=0 N) > 0 > ( + N > ( natsum ( dec N))))) > > Like that? > Am 10.02.2017 15:28 schrieb "Mike Pechkin" <mike.pech...@gmail.com>: > >> hi, >> >> On Fri, Feb 10, 2017 at 3:07 PM, Christopher Howard < >> christopher.how...@qlfiles.net> wrote: >> >> > Hi list. When I try to do >> > >> > (apply '+ (range 1 1000000) >> > >> >> >> List of millions of items is not a problem. >> Problem how you use it. >> (apply) is not for free, It *creates* a function call with a million of >> arguments. >> Stack size make sense. >> >> >> > I get segfault. I thought maybe this was some kind of internal >> > limitation of the apply function, so I defined a foldl: >> > >> > (de foldl (Fn Acc Lst) >> > (if (== () Lst) Acc >> > (let Acc2 (Fn Acc (car Lst)) >> > (foldl Fn Acc2 (cdr Lst)) ) ) ) >> > >> > : (foldl '+ 0 (range 1 1000)) >> > (foldl '+ 0 (range 1 1000)) >> > -> 500500 >> > : (foldl '+ 0 (range 1 1000000)) >> > (foldl '+ 0 (range 1 1000000)) >> > >> > >> >> Stack size makes sense too. >> I like recursions this way: >> >> (de rec1 (F A L) >> (if L >> (rec1 F (inc 'A (++ L)) L) >> A ) ) >> >> recursion with hidden accumulator from outer call and car-cdr: >> (de rec2 (F L A) >> (default A 0) >> (if L >> (rec2 F (cdr L) (inc 'A (car L))) >> A ) ) >> The call will be: (rec2 '+ (range 1 1000000))) >> >> >> As recursion just loop you can implement it without big stack too: >> (de sum1 (L) >> (let N 0 >> (for I L >> (inc 'N I) ) ) ) >> >> even without list creation: >> (de sum2 (X) >> (let N 0 >> (for I X >> (inc 'N I) ) ) ) >> >> >> And finally, that's why (sum) was implemented: >> (sum prog (range 1 1000000))) >> >> >> Bonus: for practice write recursion function to sum numbers without >> (range) > >
