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partial fix for #H5216 - sortloot inconsistency

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The tie-breaker for the qsort comparison routine used for 'sortloot'
had its logic backwards.  Instead of retaining the original relative
order for tied items, it inadvertently reversed them.  So a chest
containing
  club
  dagger named A
  dagger named B
  dagger named C
  long sword
would alternate between that order and
  club
  dagger named C
  dagger named B
  dagger named A
  long sword
each time the chest's contents were examined.  This fixes that, and
also simplifies the unnecessarily convoluted bless/curse state handling.

The other half of the report was a request that 'sortloot:n' not do
any sorting.  Right now, if the player has 'sortpack' set then
'sortloot:n' results in grouping into object classes within pack order
rather than not sorting at all.  (Also, armor and weapons are further
ordered within their groups:  armor by slot [helms, gloves, shields, &c]
and weapons by function [ammo, launchers, missiles, 'ordinary', pole-
arms].)  I think the proper fix is to add a new setting for 'sortpack'
which yields the current behavior before changing 'n' to leave things
in their unsorted order.
This commit is contained in:
PatR
2017-03-30 16:12:21 -07:00
parent 6ba906b234
commit b708f8b39b
1 changed files with 4 additions and 11 deletions
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15
src/invent.c
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@@ -166,16 +166,9 @@ const genericptr vptr2;
if ((namcmp = strcmpi(nam1, nam2)) != 0)
return namcmp;
/* Sort by BUCX. Map blessed to 4, uncursed to 2, cursed to 1, and
unknown to 0. */
val1 = obj1->bknown
? (obj1->blessed << 2)
+ ((!obj1->blessed && !obj1->cursed) << 1) + obj1->cursed
: 0;
val2 = obj2->bknown
? (obj2->blessed << 2)
+ ((!obj2->blessed && !obj2->cursed) << 1) + obj2->cursed
: 0;
/* Sort by BUCX. */
val1 = obj1->bknown ? (obj1->blessed ? 3 : !obj1->cursed ? 2 : 1) : 0;
val2 = obj2->bknown ? (obj2->blessed ? 3 : !obj2->cursed ? 2 : 1) : 0;
if (val1 != val2)
return val2 - val1; /* bigger is better */
@@ -216,7 +209,7 @@ tiebreak:
/* They're identical, as far as we're concerned. We want
to force a deterministic order, and do so by producing a
stable sort: maintain the original order of equal items. */
return (sli2->indx - sli1->indx);
return (sli1->indx - sli2->indx);
}
void