On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
I was thinking to a general *interleave()* algorithm for any
compatible Range of Range but I can't find any smart way to
process each sub range by front
Can you show a
On Wed, 23 Sep 2015 21:17:27 +, BBasile wrote:
> On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear wrote:
>> On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
>>
>>> I was thinking to a general *interleave()* algorithm for any
>>> compatible Range of Range but I can't find any
On Wednesday, 23 September 2015 at 21:17:29 UTC, BBasile wrote:
On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
I was thinking to a general *interleave()* algorithm for any
compatible Range of Range but I can't find any
On Wednesday, 23 September 2015 at 21:24:22 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 21:17:27 +, BBasile wrote:
On Wednesday, 23 September 2015 at 21:04:44 UTC, Justin Whear
wrote:
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
I was thinking to a general *interleave()* algorithm
I was thinking to a general *interleave()* algorithm for any
compatible Range of Range but I can't find any smart way to
process each sub range by front, eg:
---
void interleave(RoR)(RoR r)
{
r.each!(a => a.writeln);
}
void main()
{
auto r = [[0,2],[1,3]];
interleave(r);
}
---
On Wed, 23 Sep 2015 20:48:03 +, BBasile wrote:
> I was thinking to a general *interleave()* algorithm for any compatible
> Range of Range but I can't find any smart way to process each sub range
> by front
Can you show a sample input and output to clarify what you mean by
interleave? It's
On Wednesday, 23 September 2015 at 21:30:37 UTC, BBasile wrote:
auto interleave(RoR)(RoR r)
{
return r.transposed.join;
If you use joiner it will even be lazy and avoid the allocation.
