• Taylor Hillegeist (28/28) Feb 12 2016 So I have this code and I have to add the element
• Messenger (4/23) Feb 12 2016 Have you tried .array? I *think* it's the commonly used way to
• Nicholas Wilson (5/12) Feb 12 2016 If you need the value that a range returns (i.e. the range
• Xinok (5/7) Feb 12 2016 It might work in this case, but in general this won't work for
• Xinok (6/13) Feb 12 2016 The following combination might work:
• cym13 (2/16) Feb 12 2016 Why not just .each; ?
• Xinok (5/25) Feb 12 2016 The thing he's trying to iterate over is a range of ranges. A

Taylor Hillegeist <taylorh140 gmail.com> writes:

So I have this code and I have to add the element
.each!(a => a.each!("a"));
to the end in order for it to evaluate the range completely and 
act like I expect it too. Is there a  better thing to put in the 
place of
.each!(a => a.each!("a"));?

import std.stdio;
import std.path;
import std.file;
import std.uni;
import std.range;
import std.conv;
import std.algorithm;

void main(string[] Args){

     assert(Args.length>1,"Need a path to source files");
     assert(Args[1].isValidPath,"Path given is not Valid!");

     dirEntries(Args[1], SpanMode.depth)
         .filter!(f => f.name.endsWith(".c",".h"))
         .tee!(a => writeln("\n",a,"\n\t","=".repeat(80).join))
         .map!(a => a
             .File("r")
             .byLine
             .enumerate
             .filter!( l => l.value.byGrapheme.walkLength > 80)
             .tee!(a => writeln("Line: ",a.index,"\t",a.value))
          ).each!(a => a.each!("a")); //Force evaluation of every 
item

}

Messenger <dont shoot.me> writes:

On Friday, 12 February 2016 at 20:43:24 UTC, Taylor Hillegeist 
wrote:
 So I have this code and I have to add the element
 .each!(a => a.each!("a"));
 to the end in order for it to evaluate the range completely and 
 act like I expect it too. Is there a  better thing to put in 
 the place of
 .each!(a => a.each!("a"));?

 [...]

     dirEntries(Args[1], SpanMode.depth)
         .filter!(f => f.name.endsWith(".c",".h"))
         .tee!(a => writeln("\n",a,"\n\t","=".repeat(80).join))
         .map!(a => a
             .File("r")
             .byLine
             .enumerate
             .filter!( l => l.value.byGrapheme.walkLength > 80)
             .tee!(a => writeln("Line: ",a.index,"\t",a.value))
          ).each!(a => a.each!("a")); //Force evaluation of 
 every item

 }

Have you tried .array? I *think* it's the commonly used way to 
flatten a lazy range.

Nicholas Wilson <iamthewilsonator hotmail.com> writes:

On Friday, 12 February 2016 at 20:43:24 UTC, Taylor Hillegeist 
wrote:
 So I have this code and I have to add the element
 .each!(a => a.each!("a"));
 to the end in order for it to evaluate the range completely and 
 act like I expect it too. Is there a  better thing to put in 
 the place of
 .each!(a => a.each!("a"));?

 [...]

If you need the value that a range returns (i.e. the range 
performs "computation") then use .array

If you just want the range evaluated use walkLength

Xinok <xinok live.com> writes:

On Saturday, 13 February 2016 at 01:11:53 UTC, Nicholas Wilson 
wrote:
 ...

 If you just want the range evaluated use walkLength

It might work in this case, but in general this won't work for 
any range which defines .length as a member. In that case, 
walkLength will simply return .length of that range.

Xinok <xinok live.com> writes:

On Friday, 12 February 2016 at 20:43:24 UTC, Taylor Hillegeist 
wrote:
 So I have this code and I have to add the element
 .each!(a => a.each!("a"));
 to the end in order for it to evaluate the range completely and 
 act like I expect it too. Is there a  better thing to put in 
 the place of
 .each!(a => a.each!("a"));?

 ...

The following combination might work:

.joiner.each;

cym13 <cpicard openmailbox.org> writes:

On Saturday, 13 February 2016 at 02:17:17 UTC, Xinok wrote:
 On Friday, 12 February 2016 at 20:43:24 UTC, Taylor Hillegeist 
 wrote:
 So I have this code and I have to add the element
 .each!(a => a.each!("a"));
 to the end in order for it to evaluate the range completely 
 and act like I expect it too. Is there a  better thing to put 
 in the place of
 .each!(a => a.each!("a"));?

 ...

 The following combination might work:

 .joiner.each;





Why not just  .each;   ?

Xinok <xinok live.com> writes:

On Saturday, 13 February 2016 at 03:16:09 UTC, cym13 wrote:
 On Saturday, 13 February 2016 at 02:17:17 UTC, Xinok wrote:
 On Friday, 12 February 2016 at 20:43:24 UTC, Taylor Hillegeist 
 wrote:
 So I have this code and I have to add the element
 .each!(a => a.each!("a"));
 to the end in order for it to evaluate the range completely 
 and act like I expect it too. Is there a  better thing to put 
 in the place of
 .each!(a => a.each!("a"));?

 ...

 The following combination might work:

 .joiner.each;





 Why not just  .each;   ?

The thing he's trying to iterate over is a range of ranges. A 
single .each will only iterate over the outermost range so you 
need .joiner first to "flatten" the range, then you can use .each 
on that result.