How to force evaluation of range?
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
}
Re: How to force evaluation of range?
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;
http://dlang.org/phobos/std_algorithm_iteration.html#.joiner
http://dlang.org/phobos/std_algorithm_iteration.html#.each
Re: How to force evaluation of range?
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.
Re: How to force evaluation of range?
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
Re: How to force evaluation of range?
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.
Re: How to force evaluation of range?
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;
http://dlang.org/phobos/std_algorithm_iteration.html#.joiner
http://dlang.org/phobos/std_algorithm_iteration.html#.each
Why not just .each; ?
Re: How to force evaluation of range?
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;
http://dlang.org/phobos/std_algorithm_iteration.html#.joiner
http://dlang.org/phobos/std_algorithm_iteration.html#.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.
