如何迭代增压属性树?
-
14-10-2019 - |
题
我知道正在促进属性树的接近,并发现它是C ++编程的Boost Libs的一个很好的功能。
好吧,我有疑问吗?如何使用迭代器或类似的方式迭代属性树?
参考文献中只是一个示例,即通过:
BOOST_FOREACH
但是只有更多吗?类似于STL的容器?谈到代码质量。
解决方案
BOOST_FOREACH只是迭代的一种方便方法,可以通过迭代器来完成,begin()和end()
Your_tree_type::const_iterator end = tree.end();
for (your_tree_type::const_iterator it = tree.begin(); it != end; ++it)
...
在C ++ 11中,它是:
for (auto it: tree)
...
其他提示
这是我经过大量实验后想到的。我想在社区中分享它,因为我找不到想要的东西。每个人似乎都只是在提升文档中发布答案,我发现这是不够的。无论如何:
#include <boost/property_tree/ptree.hpp>
#include <boost/property_tree/json_parser.hpp>
#include <string>
#include <iostream>
using namespace std;
using boost::property_tree::ptree;
string indent(int level) {
string s;
for (int i=0; i<level; i++) s += " ";
return s;
}
void printTree (ptree &pt, int level) {
if (pt.empty()) {
cerr << "\""<< pt.data()<< "\"";
}
else {
if (level) cerr << endl;
cerr << indent(level) << "{" << endl;
for (ptree::iterator pos = pt.begin(); pos != pt.end();) {
cerr << indent(level+1) << "\"" << pos->first << "\": ";
printTree(pos->second, level + 1);
++pos;
if (pos != pt.end()) {
cerr << ",";
}
cerr << endl;
}
cerr << indent(level) << " }";
}
return;
}
int main(int, char*[]) {
// first, make a json file:
string tagfile = "testing2.pt";
ptree pt1;
pt1.put("object1.type","ASCII");
pt1.put("object2.type","INT64");
pt1.put("object3.type","DOUBLE");
pt1.put("object1.value","one");
pt1.put("object2.value","2");
pt1.put("object3.value","3.0");
write_json(tagfile, pt1);
ptree pt;
bool success = true;
try {
read_json(tagfile, pt);
printTree(pt, 0);
cerr << endl;
}catch(const json_parser_error &jpe){
//do error handling
success = false
}
return success;
}
这是输出:
rcook@rzbeast (blockbuster): a.out
{
"object1":
{
"type": "ASCII",
"value": "one"
},
"object2":
{
"type": "INT64",
"value": "2"
},
"object3":
{
"type": "DOUBLE",
"value": "3.0"
}
}
rcook@rzbeast (blockbuster): cat testing2.pt
{
"object1":
{
"type": "ASCII",
"value": "one"
},
"object2":
{
"type": "INT64",
"value": "2"
},
"object3":
{
"type": "DOUBLE",
"value": "3.0"
}
}
我最近遇到了这个问题,发现我的需求不完整的答案,所以我想出了这个简短而甜美的片段:
using boost::property_tree::ptree;
void parse_tree(const ptree& pt, std::string key)
{
std::string nkey;
if (!key.empty())
{
// The full-key/value pair for this node is
// key / pt.data()
// So do with it what you need
nkey = key + "."; // More work is involved if you use a different path separator
}
ptree::const_iterator end = pt.end();
for (ptree::const_iterator it = pt.begin(); it != end; ++it)
{
parse_tree(it->second, nkey + it->first);
}
}
重要的是要注意的是,除根节点外,任何节点都可以包含数据和子节点。这 if (!key.empty())
BIT将获得除根节点以外的所有数据,我们还可以开始为节点的孩子循环构建路径。
您会通过打电话开始解析 parse_tree(root_node, "")
当然,您需要在此功能内部做一些事情,以使其值得做。
如果您在不需要完整路径的地方进行一些解析,只需删除 nkey
可变及其操作,只通过 it->first
递归功能。
答案的补充 如何迭代增压属性树? :
在C ++ 11样式范围内 for (auto node : tree)
, , 每个 node
是一个 std::pair<key_type, property_tree>
而在手动书面迭代中
Your_tree_type::const_iterator end = tree.end();
for (your_tree_type::const_iterator it = tree.begin(); it != end; ++it)
...
迭代器 it
是指向这样一对的指针。用法上的差异很小。例如,要访问钥匙,一个人会写 it->first
但 node.first
.
作为新答案发布,因为我对原始答案的拟定编辑被拒绝,并提出了发布新答案的建议。
如果我们想执行一些算法操纵,则可以使用基于BFS的PRING PTREE遍历
int print_ptree_bfs(ptree &tree) {
try {
std::queue<ptree*> treeQ;
std::queue<string> strQ;
ptree* temp;
if (tree.empty())
cout << "\"" << tree.data() << "\"";
treeQ.push(&tree);
//cout << tree.data();
strQ.push(tree.data());
while (!treeQ.empty()) {
temp = treeQ.front();
treeQ.pop();
if (temp == NULL) {
cout << "Some thing is wrong" << std::endl;
break;
}
cout << "----- " << strQ.front() << "----- " << std::endl;
strQ.pop();
for (auto itr = temp->begin(); itr != temp->end(); itr++) {
if (!itr->second.empty()) {
//cout << itr->first << std::endl;
treeQ.push(&itr->second);
strQ.push(itr->first);
} else {
cout<<itr->first << " " << itr->second.data() << std::endl;
}
}
cout << std::endl;
}
} catch (std::exception const& ex) {
cout << ex.what() << std::endl;
}
return EXIT_SUCCESS;
}