本文整理汇总了C++中NodeSet::subtreesAreDisjoint方法的典型用法代码示例。如果您正苦于以下问题：C++ NodeSet::subtreesAreDisjoint方法的具体用法？C++ NodeSet::subtreesAreDisjoint怎么用？C++ NodeSet::subtreesAreDisjoint使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类NodeSet的用法示例。
在下文中一共展示了NodeSet::subtreesAreDisjoint方法的2个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: evaluate
void LocationPath::evaluate(NodeSet& nodes) const
{
bool resultIsSorted = nodes.isSorted();
for (unsigned i = 0; i < m_steps.size(); i++) {
Step* step = m_steps[i];
NodeSet newNodes;
HashSet<Node*> newNodesSet;
bool needToCheckForDuplicateNodes = !nodes.subtreesAreDisjoint() || (step->axis() != Step::ChildAxis && step->axis() != Step::SelfAxis
&& step->axis() != Step::DescendantAxis && step->axis() != Step::DescendantOrSelfAxis && step->axis() != Step::AttributeAxis);
if (needToCheckForDuplicateNodes)
resultIsSorted = false;
// This is a simplified check that can be improved to handle more cases.
if (nodes.subtreesAreDisjoint() && (step->axis() == Step::ChildAxis || step->axis() == Step::SelfAxis))
newNodes.markSubtreesDisjoint(true);
for (unsigned j = 0; j < nodes.size(); j++) {
NodeSet matches;
step->evaluate(nodes[j], matches);
if (!matches.isSorted())
resultIsSorted = false;
for (size_t nodeIndex = 0; nodeIndex < matches.size(); ++nodeIndex) {
Node* node = matches[nodeIndex];
if (!needToCheckForDuplicateNodes || newNodesSet.add(node).isNewEntry)
newNodes.append(node);
}
}
nodes.swap(newNodes);
}
nodes.markSorted(resultIsSorted);
}示例2: evaluate
void LocationPath::evaluate(NodeSet& nodes) const
{
bool resultIsSorted = nodes.isSorted();
for (auto& step : m_steps) {
NodeSet newNodes;
HashSet<Node*> newNodesSet;
bool needToCheckForDuplicateNodes = !nodes.subtreesAreDisjoint() || (step->axis() != Step::ChildAxis && step->axis() != Step::SelfAxis
&& step->axis() != Step::DescendantAxis && step->axis() != Step::DescendantOrSelfAxis && step->axis() != Step::AttributeAxis);
if (needToCheckForDuplicateNodes)
resultIsSorted = false;
// This is a simplified check that can be improved to handle more cases.
if (nodes.subtreesAreDisjoint() && (step->axis() == Step::ChildAxis || step->axis() == Step::SelfAxis))
newNodes.markSubtreesDisjoint(true);
for (auto& node : nodes) {
NodeSet matches;
step->evaluate(*node, matches);
if (!matches.isSorted())
resultIsSorted = false;
for (auto& match : matches) {
if (!needToCheckForDuplicateNodes || newNodesSet.add(match.get()).isNewEntry)
newNodes.append(match.copyRef());
}
}
nodes = WTFMove(newNodes);
}
nodes.markSorted(resultIsSorted);
}