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XPath: Create set for a|b in order before duplicate filtering

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This does not change the result of a union operation [substantially], but
it means that we now give a list to remove_duplicates that has more natural
ordering.

If remove_duplicates didn't sort the array, we'd have union operations
resulting in a consistent predictable order.

Contributes to #254.
This commit is contained in:
Arseny Kapoulkine 2019-02-26 21:35:05 -08:00
parent d902709da2
commit 93c7bacb29
1 changed files with 6 additions and 6 deletions
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12
src/pugixml.cpp
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@ -10737,16 +10737,16 @@ PUGI__NS_BEGIN
xpath_stack swapped_stack = {stack.temp, stack.result};
xpath_node_set_raw ls = _left->eval_node_set(c, swapped_stack, eval);
xpath_node_set_raw rs = _right->eval_node_set(c, stack, eval);
xpath_node_set_raw ls = _left->eval_node_set(c, stack, eval);
xpath_node_set_raw rs = _right->eval_node_set(c, swapped_stack, eval);
// we can optimize merging two sorted sets, but this is a very rare operation, so don't bother
rs.set_type(xpath_node_set::type_unsorted);
ls.set_type(xpath_node_set::type_unsorted);
rs.append(ls.begin(), ls.end(), stack.result);
rs.remove_duplicates();
ls.append(rs.begin(), rs.end(), stack.result);
ls.remove_duplicates();
return rs;
return ls;
}
case ast_filter: