• "Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> (9/9) Jan 22 2014 import std.algorithm;
• monarch_dodra (5/14) Jan 22 2014 Because "remove" takes an "offset" as an argument, not an element.
• "Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> (4/8) Jan 22 2014 Thanks a lot. I was trying to get that part to work, but I had a
• bearophile (4/7) Jan 22 2014 is that removing only 0 or 1 items?
• "Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> (2/9) Jan 22 2014 It removes all items that match the tid.
• monarch_dodra (3/10) Jan 22 2014 Maybe you confusing the new style lambda for a "greater equal"
• bearophile (6/8) Jan 22 2014 My point was that the shown code doesn't remove only one item in
• monarch_dodra (33/42) Jan 22 2014 Ah... I see. Yeah, this will remove *all* items that match the
• bearophile (8/9) Jan 22 2014 There will be.
• monarch_dodra (6/15) Jan 22 2014 Hum... that requires iterating the range twice for a non-RA

"Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> writes:

import std.algorithm;
import std.concurrency;

void main() {
	Tid[] tids = [];
	Tid tid = thisTid;
	tids ~= tid;
	tids.remove(tid);
}

Why does this not compile?

"monarch_dodra" <monarchdodra gmail.com> writes:

On Wednesday, 22 January 2014 at 12:11:22 UTC, Casper Færgemand 
wrote:
 import std.algorithm;
 import std.concurrency;

 void main() {
 	Tid[] tids = [];
 	Tid tid = thisTid;
 	tids ~= tid;
 	tids.remove(tid);
 }

 Why does this not compile?

Because "remove" takes an "offset" as an argument, not an element.

To remove an element, I *think* you do it this way:

tids = tids.remove!(a=>a == tid)();

"Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> writes:

On Wednesday, 22 January 2014 at 13:16:18 UTC, monarch_dodra 
wrote:
 Because "remove" takes an "offset" as an argument, not an 
 element.

 To remove an element, I *think* you do it this way:

 tids = tids.remove!(a=>a == tid)();

Thanks a lot. I was trying to get that part to work, but I had a 
hard time realizing that => was a lambda and not a >=. x.x

"bearophile" <bearophileHUGS lycos.com> writes:

Casper Færgemand:

 To remove an element, I *think* you do it this way:

 tids = tids.remove!(a=>a == tid)();


is that removing only 0 or 1 items?

Bye,
bearophile

"Casper =?UTF-8?B?RsOmcmdlbWFuZCI=?= <shorttail hotmail.com> writes:

On Wednesday, 22 January 2014 at 13:51:51 UTC, bearophile wrote:
 Casper Færgemand:

 To remove an element, I *think* you do it this way:

 tids = tids.remove!(a=>a == tid)();


 is that removing only 0 or 1 items?

 Bye,
 bearophile

It removes all items that match the tid.

"monarch_dodra" <monarchdodra gmail.com> writes:

On Wednesday, 22 January 2014 at 13:51:51 UTC, bearophile wrote:
 Casper Færgemand:

 To remove an element, I *think* you do it this way:

 tids = tids.remove!(a=>a == tid)();


 is that removing only 0 or 1 items?

 Bye,
 bearophile

Maybe you confusing the new style lambda for a "greater equal" 
operator? I can't make sense of your question any other way.

"bearophile" <bearophileHUGS lycos.com> writes:

monarch_dodra:

 Maybe you confusing the new style lambda for a "greater equal" 
 operator? I can't make sense of your question any other way.

My point was that the shown code doesn't remove only one item in 
presence of duplicated ones. In this case tid are unique, but in 
general using that code to remove one item is not a good idea.

Bye,
bearophile

"monarch_dodra" <monarchdodra gmail.com> writes:

On Wednesday, 22 January 2014 at 15:41:58 UTC, bearophile wrote:
 monarch_dodra:

 Maybe you confusing the new style lambda for a "greater equal" 
 operator? I can't make sense of your question any other way.

 My point was that the shown code doesn't remove only one item 
 in presence of duplicated ones. In this case tid are unique, 
 but in general using that code to remove one item is not a good 
 idea.

 Bye,
 bearophile

Ah... I see. Yeah, this will remove *all* items that match the 
TID. I'm not sure that's a problem in this context, but I you did 
want to remove "at most" 1 item, then this isn't the correct 
solution.

There's no phobos solution for that, but I guess it would be 
written something like:

template removeOne(alias pred, SwapStrategy s = 
SwapStrategy.stable)
{
     Range removeOne(Range)(Range range)
     {
         auto result = range.save;
         auto f = find!pred(range);
         if (f.empty) return result;

         static if (s == SwapStrategy.stable)
         {
             auto ff = f.save;
             f.popFront();
             ff.popBack;
             for ( ; !f.empty; f.popFront(), ff.popFront())
                 moveFront(f, ff);
         }
         else
         {
             move(find.back, find.front);
         }
         result.popBack();
         return result;
     }
}

Disclaimer: Not actually tested. May also horribly fail on 
non-reference ranges.

"bearophile" <bearophileHUGS lycos.com> writes:

monarch_dodra:

 There's no phobos solution for that,

There will be.

In the meantime use:
items = items.remove(items.countUntil(needle));

See also:
https://d.puremagic.com/issues/show_bug.cgi?id=10959

Bye,
bearophile

"monarch_dodra" <monarchdodra gmail.com> writes:

On Wednesday, 22 January 2014 at 16:48:45 UTC, bearophile wrote:
 monarch_dodra:

 There's no phobos solution for that,

 There will be.

 In the meantime use:
 items = items.remove(items.countUntil(needle));

Hum... that requires iterating the range twice for a non-RA 
range. And you forgot a save:

items = items.remove(items.save.countUntil(needle));

But it *is* much simpler.

 See also:
 https://d.puremagic.com/issues/show_bug.cgi?id=10959

Thx.

 Bye,
 bearophile