On Saturday 12 March 2011 02:48:19 Russel Winder wrote:
> Jonathan,
>
> On Sat, 2011-03-12 at 10:31 +, 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
> > > passi
Jonathan,
On Sat, 2011-03-12 at 10:31 +, 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 con
On Saturday 12 March 2011 02:31:20 Russel Winder wrote:
> Jonathan,
>
> Thanks for the info, very helpful. One point though:
>
> On Sat, 2011-03-12 at 01:56 -0800, Jonathan M Davis wrote:
> [ . . . ]
>
> > What's happening is that the parameter that you're passing n to for
> > recurrence is siz
On Sat, 2011-03-12 at 02:15 -0800, Ali Çehreli wrote:
[ . . . ]
> void main()
> {
> long[] data = [ 0, 1, 1, 2, 3, 5, 8 ];
>
> foreach (n; 0 .. data.length) {
> assert(equal(declarative(n), data[0..n]));
> }
> }
In fact the driver is:
unittest {
immutable data = [
[
Jonathan,
Thanks for the info, very helpful. One point though:
On Sat, 2011-03-12 at 01:56 -0800, Jonathan M Davis 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
>
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]
On Saturday 12 March 2011 01:33:34 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
On Sat, 2011-03-12 at 09:33 +, Russel Winder wrote:
[ . . . ]
> Interestingly, or not, the code:
>
> long declarative ( immutable long n ) {
> return take ( recurrence ! ( "a[n-1] + a[n-2]" ) ( 0L , 1L ) , n ) ;
> }
>
> results in the return statement delivering:
>
> rdmd --main -unittest
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)
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);
> }
Just curious: How can I get ulong here?
foreach (f; take(recurrence!("a[n-1] + a[n-2]")(0, 1), 50))
{
writeln(f);
}
Results:
0
1
1
2
3
5
8
..
1134903170
1836311903
-1323752223
512559680
-811192543
11 matches
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