On Saturday, 11 April 2015 at 10:53:46 UTC, Jakob Ovrum wrote:
On Saturday, 11 April 2015 at 10:50:17 UTC, matovitch wrote:
Hello,

The question is in the title. It should be possible for a finite random access ranges to perform an indexed foreach no ? I mean like :

foreach(size_t i = 0, auto ref x; R)
{
   /*...*/
}

Why are other foreach statements overloadable but this one ?

Thanks in advance.

As of 2.067, you can use std.range.enumerate[1]. See the PR that added it[2] and the enhancement request that proposed it[3] for more information about why it's a library function.

[1] http://dlang.org/phobos/std_range#enumerate
[2] https://github.com/D-Programming-Language/phobos/pull/1866
[3] https://issues.dlang.org/show_bug.cgi?id=5550

Thanks for the tip...I just tried it on the generic kmeans algorithm I coded, there are huge performance issue with dmd 2.0.67 :

//With foreach ennumerate
cbrugel@eleanor ~/w/D/kmeans++> time ./kmeans_example
Point(0.742677, 0.749284, 0.746855)
Point(0.246975, 0.247246, 0.251123)
Point(0.751372, 0.754126, 0.247526)
Point(0.250743, 0.754682, 0.250682)
Point(0.755332, 0.249898, 0.749533)
Point(0.254945, 0.25063, 0.750403)
Point(0.746505, 0.258751, 0.249303)
Point(0.244185, 0.748149, 0.750536)
4.72user 0.00system 0:04.73elapsed 99%CPU (0avgtext+0avgdata 4796maxresident)k
0inputs+0outputs (0major+697minor)pagefaults 0swaps

//with a classic for loop
cbrugel@eleanor ~/w/D/kmeans++> dmd kmeans_example.d kmeans.d
cbrugel@eleanor ~/w/D/kmeans++> time ./kmeans_example
Point(0.744609, 0.251452, 0.252298)
Point(0.750793, 0.754791, 0.248945)
Point(0.752109, 0.245865, 0.754593)
Point(0.752743, 0.746093, 0.748006)
Point(0.250339, 0.749277, 0.746064)
Point(0.249227, 0.24674, 0.751623)
Point(0.250478, 0.745153, 0.245349)
Point(0.243387, 0.249, 0.24996)
1.30user 0.00system 0:01.30elapsed 99%CPU (0avgtext+0avgdata 6048maxresident)k
0inputs+0outputs (0major+560minor)pagefaults 0swaps

How does ennumerate work does it provide an other methods the other range don't and that is used by the indexed foreach ?

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