• Andrew Edwards (23/23) Aug 04 2014 Is there a way to take a bounded rage from a infinite forward range?
• Brad Anderson (10/35) Aug 04 2014 I'd use std.algorithm.until:
• Andrew Edwards (2/5) Aug 04 2014 Precisely what I was looking for. Thanks.

Andrew Edwards <ridimz yahoo.com> writes:

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?

"Brad Anderson" <eco gnuk.net> writes:

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);
    }

Andrew Edwards <ridimz yahoo.com> writes:

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.