• ARaspiK (3/3) Feb 25 2018 Instead of passing std.range.zip a set of ranges as different
• Paul Backus (11/14) Feb 25 2018 `std.range.transposed` does this, but it requires that the range
• ARaspiK (3/17) Feb 25 2018 Thank you so much. It works now. I was already receiving a

ARaspiK <arav1025 gmail.com> writes:

Instead of passing std.range.zip a set of ranges as different 
arguments, is it possible to hand the m a range of ranges, and 
get them to zip together each element of every subrange?

Paul Backus <snarwin gmail.com> writes:

On Sunday, 25 February 2018 at 16:22:19 UTC, ARaspiK wrote:
 Instead of passing std.range.zip a set of ranges as different 
 arguments, is it possible to hand the m a range of ranges, and 
 get them to zip together each element of every subrange?

`std.range.transposed` does this, but it requires that the range 
of ranges has assignable elements, so it may not work in all 
cases. For example:

import std.range;
import std.algorithm.iteration;
import std.stdio;

auto rr1 = [[1, 2, 3], [4, 5, 6]];
rr1.transposed.each!writeln; // Works

auto rr2 = only(only(1, 2, 3), only(4, 5, 6));
rr2.transposed.each!writeln; // Doesn't work

ARaspiK <arav1025 gmail.com> writes:

On Sunday, 25 February 2018 at 20:18:27 UTC, Paul Backus wrote:
 On Sunday, 25 February 2018 at 16:22:19 UTC, ARaspiK wrote:
 Instead of passing std.range.zip a set of ranges as different 
 arguments, is it possible to hand the m a range of ranges, and 
 get them to zip together each element of every subrange?

 `std.range.transposed` does this, but it requires that the 
 range of ranges has assignable elements, so it may not work in 
 all cases. For example:

 import std.range;
 import std.algorithm.iteration;
 import std.stdio;

 auto rr1 = [[1, 2, 3], [4, 5, 6]];
 rr1.transposed.each!writeln; // Works

 auto rr2 = only(only(1, 2, 3), only(4, 5, 6));
 rr2.transposed.each!writeln; // Doesn't work

Thank you so much. It works now. I was already receiving a 
forward range, so copying was easy.