在函数中,我使用pugi首先加载XML文件。然后,我遍历树的子xml节点,并将一些子xml节点(xml_node类型的对象)推送到xml_node的向量。但是一旦我退出此函数,就会删除从XML文件加载的原始XML树结构对象,导致xml节点向量中的元素变为无效。
下面是一个示例代码(快速编写)以显示:
#include "pugixml.hpp"
#include <vector>
void ProcessXmlDeferred( std::vector<pugi::xml_node> const &subTrees )
{
for( auto & const node: subTrees)
{
// parse each xml_node node
}
}
void IntermedProcXml( pugi::xml_node const &node)
{
// parse node
}
std::vector<pugi::xml_node> BuildSubTrees(pugi::xml_node const & node )
{
std::vector<pugi::xml_node> subTrees;
pugi::xml_node temp = node.child("L1");
subTrees.push_back( temp );
temp = node.child.child("L2");
subTrees.push_back( temp );
temp = node.child.child.child("L3");
subTrees.push_back( temp );
return subTrees;
}
void LoadAndProcessDoc( const char* fileNameWithPath, std::vector<pugi::xml_node> & subTrees )
{
pugi::xml_document doc;
pugi::xml_parse_result result = doc.load( fileNameWithPath );
subTrees = BuildSubTrees( result.child("TOP") );
IntermedProcXml( result.child("CENTRE") );
// Local pugi objects("doc" and "result") destroyed at exit of this
// function invalidating contents of xml nodes inside vector "subTrees"
}
int main()
{
char fileName[] = "myFile.xml";
std::vector<pugi::xml_node> myXmlSubTrees;
// Load XML file and return vector of XML sub-tree's for later parsing
LoadAndProcessDoc( fileName, myXmlSubTrees );
// At this point, the contents of xml node's inside the vector
// "myXmlSubTrees" are no longer valid and thus are unsafe to use
// ....
// Lots of intermediate code
// ....
// This function tries to process vector whose xml nodes
// are invalid and thus throws errors
ProcessXmlDeferred( myXmlSubTrees );
return 0;
}
因此,我需要一种方法来保存/复制/克隆/移动原始XML树的子树(xml节点),这样我甚至可以在原始XML根树对象之后安全地解析它们。被删除。怎么在pugi做到这一点?
答案 0 :(得分:1)
只需将xml_document对象的所有权传递给调用者即可。
您可以通过强制调用者提供xml_document对象(向函数添加xml_document&
参数),或者从函数中返回shared_ptr<xml_document>
或unique_ptr<xml_document>
来执行此操作。节点向量。