(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)
