Fibonacci with ranges

[Russel Winder]

Jonathan,

On Sat, 2011-03-12 at 10:31 +0000, Russel Winder wrote:
[ . . . ]
> > What's happening is that the parameter that you're passing n to for recurrence 
> > is size_t. And on 32-bit systems, size_t is uint, so passing n - which is long - 
> > to recurrence would be a narrowing conversion, which requires a cast. The 
> > correct thing to do would be make n a size_t. The other thing that you'd need to 
> > do is change declarative to return auto, since take returns a range, _not_ a 
> > long.

To analyse this a bit more I temporarily deconstructed the expression:

        long declarative ( immutable long n ) {
          auto r = recurrence ! ( "a[n-1] + a[n-2]" ) ( 0L , 1L ) ;
          auto t = take ( r , cast ( size_t ) ( n ) ) ;
          return t [ n ] ;
          //return ( take ( recurrence ! ( "a[n-1] + a[n-2]" ) ( 0L , 1L ) , cast ( size_t ) ( n ) ) ) [ n ] ;
        }

So with the cast it compiles fine -- though it still seems to me to be
beyond the point of comprehension as to why an applications programmer
has to manually cast a long to a size_t.  However the indexing of the
range fails:

        fibonacci_d2.d(17): Error: no [] operator overload for type Take!(Recurrence!(fun,long,2u))

Which elicits the response:  for f$$$$ sake, I'm just copying the
example from the manual.

OK, so I am grumpy this morning, but that doesn't affect the fact that
there appears to be a disconnect between documentation and what actually
works.
-- 
Russel.
=============================================================================
Dr Russel Winder      t: +44 20 7585 2200   voip: sip:russel.winder at ekiga.net
41 Buckmaster Road    m: +44 7770 465 077   xmpp: russel at russel.org.uk
London SW11 1EN, UK   w: www.russel.org.uk  skype: russel_winder
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