我有一个带有数字的树和一个布尔值来查看这些数字是否已被求和这是代码：
#include <iostream>
#include <string>
#include <boost/thread.hpp>
#include <boost/lambda/bind.hpp>
using namespace std;
class ThreadBase
{
private:
boost::shared_ptr<boost::thread> m_thread_ptr;
public:
ThreadBase() : m_thread_ptr() { }
virtual ~ThreadBase() { }
virtual void run() = 0;
void start()
{
if (m_thread_ptr == NULL)
{
m_thread_ptr.reset(
new boost::thread(
boost::lambda::bind(&ThreadBase::run, this)));
}
else
{
throw std::runtime_error("multiple start");
}
}
void join()
{
if (m_thread_ptr)
{
m_thread_ptr->join();
}
}
};
struct NodeT
{
int key_value;
NodeT *left;
NodeT *right;
bool done;
};
class Btree
{
public:
Btree();
~Btree();
void insert(int key);
NodeT *search(int key);
void destroy_tree();
void inOrder(NodeT *leaf);
NodeT *root;
void display(NodeT *leaf, string where);
private:
void destroy_tree(NodeT *leaf);
void insert(int key, NodeT *leaf);
NodeT *search(int key, NodeT *leaf);
};
Btree::Btree()
{
cout << "creating tree" << endl;
root=NULL;
}
Btree::~Btree()
{
cout << "deleting tree end" << endl;
destroy_tree() ;
}
void Btree::destroy_tree(NodeT *leaf)
{
if (leaf != NULL)
{
destroy_tree(leaf->left);
destroy_tree(leaf->right);
delete leaf;
cout << "tree deleted " << leaf->key_value << endl;
}
}
void Btree::insert(int key, NodeT *leaf)
{
if (key < leaf->key_value)
{
if (leaf->left != NULL)
insert(key, leaf->left);
else
{
cout << "creating new node to insert with less value" << key << endl;
leaf->left = new NodeT;
leaf->left->key_value=key;
leaf->left->left=NULL; //sets the left child to null
leaf->left->right=NULL; //sets the right child to null
}
}
else if (key >=leaf->key_value)
{
if (leaf->right != NULL)
insert(key, leaf->right);
else
{
cout << "creating node with more or equal value" << key << endl;
leaf->right = new NodeT;
leaf->right->key_value=key;
leaf->right->left=NULL; //sets the left node to null
leaf->right->right=NULL; //sets the right node to nulll
}
}
}
NodeT *Btree::search(int key, NodeT *leaf)
{
cout << "searching node" << key << endl;
if (leaf != NULL)
{
if(key == leaf->key_value)
{
cout << "finded in first attempt " << endl;
return leaf;
}
if(key < leaf->key_value)
{
cout << "looking again with less value to the left" << endl;
return search(key, leaf->left);
}
else
{
cout << "looking again more value to the right" << endl;
return search(key, leaf->right);
}
}
else
{
cout << "not found" << endl;
return NULL;
}
}
void Btree::insert(int key)
{
if (root != NULL)
insert(key, root);
else
{
cout << "inserting new node private func" << key << endl;
root=new NodeT;
root->key_value=key;
root->right=NULL;
root->left=NULL;
}
}
NodeT *Btree::search(int key)
{
cout << "searhing node private func" << endl;
return search(key, root);
}
void Btree::destroy_tree()
{
cout << "deleting tree public" << endl;
destroy_tree(root);
}
void Btree::inOrder(NodeT *leaf)
{
if(leaf!= NULL)
{
inOrder(leaf->left);
cout << "search inorder" << leaf->key_value << endl;
inOrder(leaf->right);
}
}
void Btree::display(NodeT *leaf, string where)
{
cout << where << endl;
if (leaf != NULL)
{
cout << leaf->key_value << " t or f"<< leaf->done << endl;
display(leaf->left, "left");
display(leaf->right, "rigth");
}
}
class MyThread : public ThreadBase
{
public:
void run()
{
}
int sum( Btree *ptrRoot)
{
//pointer to next node or to parent
NodeT *next = ptrRoot->root;
//if we are in some root we should be here so leave
if (next->left == NULL && next->right == NULL)
{
return 0;
}
//we are in a parent with 2 nodes
else if(next->left->left != NULL && next->right->right != NULL && next->left->done == false && next->right->done == false)
{
//sum values and put it into this node and set bool to true
//to not go again
next->key_value = next->left->key_value + next->right->key_value;
next->left->done = true;
next->right->done = true;
//all good
return 0;
}
//we are in a parent with 1 node the right one
else if(next->left->left == NULL && next->right->right != NULL && next->right->done == false)
{
//sum value + 0 and set flag to true
next->key_value = next->right->key_value + 0;
next->right->done = true;
//all is good
return 0;
}
//we are in a parent with the left node
else if (next->left->left != NULL && next->right->right == NULL && next->left->done == false)
{
//sum value + 0 and set flag to true
next->key_value = next->left->key_value + 0;
next->left->done = true;
//all is good
return 0;
}
cout << "we are at here " << next->key_value << endl;
return 0;
}
};
int THREADS_HOW_MANY = 0;
int main()
{
cout << "Trees example" << endl;
cout << "make yoir choice" << endl;
cout << "1 insert 2 delete 3 search" << endl;
Btree bt;
bt.insert(10);
bt.insert(15);
bt.insert(4);
bt.insert(2);
bt.insert(8);
bt.search(4);
bt.search(2);
bt.search(20);
bt.inOrder(bt.root);
bt.display(bt.root, "root");
cout << "end everyrhing" << endl;
MyThread mt;
mt.start();
mt.run();
mt.sum(&bt);
THREADS_HOW_MANY = boost::thread::hardware_concurrency();
std::cout << THREADS_HOW_MANY << std::endl;
return 0;
}
我想要做的是去所有的树，看看有假的叶子，并在父母和父母等总和，直到根，这必须是线程，所以1线程总和前2而另一个是无限的下一个广告，现在我认为我的方式有点偏离，因为功能总和尚未完成，所以你能给出一些提示。
答案 0 :(得分：0)
您将从不同的线程到达同一节点，因此您需要在访问node.done或使用“无锁定”技术时使用锁定，例如node.done成员的比较和交换。在对节点求和之前，您需要修改代码以更改“完成”的值。