Taking from infinite forward ranges

[Brad Anderson via Digitalmars-d-learn]

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);
    }