Is there a way to take a bounded rage from a infinite forward range?

Given the Fibonacci sequence:

	auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);

I can take the first n elements:

	take(fib, 10);

But say I want all positive elements below 50000 in value (there are eight such values [2, 8, 34, 144, 610, 2584, 10946, 46368]), how would I "take" them? Of course I could filter the range, leaving only positive values, and then take(fib, 8). But what if I didn't know there were 8, how could I take them from there filtered range?

Currently I do this:

	foreach(e; fib)
	{
	    if (e >= val) break;
	    // so something with e
	}

or

	while((e = fib.front()) < n)
	{
	    // do something with e
	    fib.popFront();
	}

Is there a better way?

On Tuesday, 5 August 2014 at 01:23:19 UTC, Andrew Edwards wrote:
> Is there a way to take a bounded rage from a infinite forward range?
>
> Given the Fibonacci sequence:
>
> 	auto fib = recurrence!("a[n-1] + a[n-2]")(1, 1);
>
> I can take the first n elements:
>
> 	take(fib, 10);
>
> But say I want all positive elements below 50000 in value (there are eight such values [2, 8, 34, 144, 610, 2584, 10946, 46368]), how would I "take" them? Of course I could filter the range, leaving only positive values, and then take(fib, 8). But what if I didn't know there were 8, how could I take them from there filtered range?
>
> Currently I do this:
>
> 	foreach(e; fib)
> 	{
> 	    if (e >= val) break;
> 	    // so something with e
> 	}
>
> or
>
> 	while((e = fib.front()) < n)
> 	{
> 	    // do something with e
> 	    fib.popFront();
> 	}
>
> Is there a better way?

I'd use std.algorithm.until:

   void main()
   {
     import std.algorithm, std.range, std.stdio;

     auto fib_until_50k = recurrence!("a[n-1] + a[n-2]")(1, 1)
                         .until!(a => a > 50_000);
	
     writeln(fib_until_50k);
   }

On 8/5/14, 10:28 AM, Brad Anderson wrote:
> On Tuesday, 5 August 2014 at 01:23:19 UTC, Andrew Edwards wrote:
>> Is there a way to take a bounded rage from a infinite forward range?
>
> I'd use std.algorithm.until:
>

Precisely what I was looking for. Thanks.