Fibonacci with ranges

[Ali Çehreli]

On 03/12/2011 01:33 AM, Russel Winder wrote:
 > On Fri, 2011-03-11 at 18:46 -0500, Jesse Phillips wrote:
 >> Without testing: foreach (f; take(recurrence!("a[n-1] + 
a[n-2]")(0UL, 1UL), 50))
 >>
 >> teo Wrote:
 >>
 >>> Just curious: How can I get ulong here?
 >>>
 >>> foreach (f; take(recurrence!("a[n-1] + a[n-2]")(0, 1), 50))
 >>> {
 >>> 	writeln(f);
 >>> }
 >>
 >
 > Interestingly, or not, the code:
 >
 > long declarative ( immutable long n ) {
 >    return take ( recurrence ! ( "a[n-1] + a[n-2]" ) ( 0L , 1L ) , n ) ;
 > }

take returns a lazy range which can't be returned as a single long.

Reading your other post, I think this may be what you wanted to see:

import std.range;
import std.algorithm;

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

void main()
{
     long[] data = [ 0, 1, 1, 2, 3, 5, 8 ];

     foreach (n; 0 .. data.length) {
         assert(equal(declarative(n), data[0..n]));
     }
}

Ali