std.range lockstep is not input range but opApply entity. Workarounds?

[mist]

I basically want to be able to do stuff like this:
auto result = map!( (a, b) = a+b )( lockstep(range1, range2) );

Are there any standard short ways to wrap an input range around 
struct with opApply (which Lockstep is)?


Also what about redesigning Lockstep as a proper range? I could 
do a pull request but not sure about current intentions.

Re: std.range lockstep is not input range but opApply entity. Workarounds?

[Simen Kjaeraas]

On 2012-48-29 14:12, mist n...@none.none wrote:


I basically want to be able to do stuff like this:
auto result = map!( (a, b) = a+b )( lockstep(range1, range2) );

Are there any standard short ways to wrap an input range around struct  
with opApply (which Lockstep is)?


Also what about redesigning Lockstep as a proper range? I could do a  
pull request but not sure about current intentions.


Use std.range.zip instead:

  auto result = map!( (a, b) = a+b )( zip(range1, range2) );

The reason there are two ways is lockstep works better with foreach:

  foreach (a, b; lockstep(A, B) ) {
  // Use a and b here.
  }

Contrast with zip:

  foreach (a; zip(A, B) ) {
  // Use a[0] and a[1] here.
  }

There have been suggestions to better integrate tuples in the language,
so in the future zip may have all the advantages of lockstep (and vice
versa), but don't cross your fingers.

--
Simen

Re: std.range lockstep is not input range but opApply entity. Workarounds?

[mist]

Any objections to documentation update to cross-reference zip and 
lockstep to each other? Was not even searching for first one when 
found lockstep, huh.


On Saturday, 29 December 2012 at 14:19:39 UTC, Simen Kjaeraas 
wrote:

On 2012-48-29 14:12, mist n...@none.none wrote:


I basically want to be able to do stuff like this:
auto result = map!( (a, b) = a+b )( lockstep(range1, range2) 
);


Are there any standard short ways to wrap an input range 
around struct with opApply (which Lockstep is)?


Also what about redesigning Lockstep as a proper range? I 
could do a pull request but not sure about current intentions.


Use std.range.zip instead:

  auto result = map!( (a, b) = a+b )( zip(range1, range2) );

The reason there are two ways is lockstep works better with 
foreach:


  foreach (a, b; lockstep(A, B) ) {
  // Use a and b here.
  }

Contrast with zip:

  foreach (a; zip(A, B) ) {
  // Use a[0] and a[1] here.
  }

There have been suggestions to better integrate tuples in the 
language,
so in the future zip may have all the advantages of lockstep 
(and vice

versa), but don't cross your fingers.

Re: std.range lockstep is not input range but opApply entity. Workarounds?

[Ali Çehreli]

On 12/29/2012 06:19 AM, Simen Kjaeraas wrote:

 foreach (a; zip(A, B) ) {
 // Use a[0] and a[1] here.
 }

 There have been suggestions to better integrate tuples in the language,
 so in the future zip may have all the advantages of lockstep (and vice
 versa), but don't cross your fingers.

Tuples have already been integrated at least as much: :)

import std.stdio;
import std.range;

void main()
{
int[] cats;
int[] birds;

foreach (cat, bird; zip(cats, birds)) {
writeln(cat, bird);
}
}

Ali

Re: std.range lockstep is not input range but opApply entity. Workarounds?

[mist]

Not clever enough to expand like this though:
map!( (a, b) = a+b )( zip(Range1, Range2) );

Using a = a[0]+a[1] is not that big deal though.

On Saturday, 29 December 2012 at 17:58:55 UTC, Ali Çehreli wrote:

On 12/29/2012 06:19 AM, Simen Kjaeraas wrote:

 foreach (a; zip(A, B) ) {
 // Use a[0] and a[1] here.
 }

 There have been suggestions to better integrate tuples in the
language,
 so in the future zip may have all the advantages of lockstep
(and vice
 versa), but don't cross your fingers.

Tuples have already been integrated at least as much: :)

import std.stdio;
import std.range;

void main()
{
int[] cats;
int[] birds;

foreach (cat, bird; zip(cats, birds)) {
writeln(cat, bird);
}
}

Ali

Re: std.range lockstep is not input range but opApply entity. Workarounds?

[Simen Kjaeraas]

On 2012-52-29 20:12, mist n...@none.none wrote:


Not clever enough to expand like this though:
map!( (a, b) = a+b )( zip(Range1, Range2) );

Using a = a[0]+a[1] is not that big deal though.


That oughta be doable. However, seeing as std.functional only contains
unaryFun and binaryFun (dranges has naryFun), this approach cannot
currently extend to tuples with more than two fields.

--
Simen