返回的指针属性中类实例的矢量属性消失

Question

我正在实现一种树搜索，该树搜索需要能够从树中获取“最有希望的节点”，然后对该节点执行某些操作，以便为下一次迭代更新树的其余部分。

问题：对象指针Board*的向量属性似乎在生成它们的函数的return和在调用环境中将其保存的Board*值之间改变。

我的输出：

>>g++ -std=c++17 -o tmp.out tests/test.cpp // <- require c++17 for other parts of the project
>>./tmp.out 
Best leaf by tree traversal has score: 8
Best leaf associated state has -1977735524 values in its attribute vector though! (Should be 4)
我期望的是：

>>g++ -std=c++17 -o tmp.out tests/test.cpp // <- require c++17 for other parts of the project
>>./tmp.out 
Best leaf by tree traversal has score: 8
Best leaf associated state has 4 values in its attribute vector though! (Should be 4)
#include <iostream>
#include <vector>
#include <queue>

using namespace std;

class Board{
    vector<int> attribute;
    string name;
public:
    Board(){
        attribute = {1,2,3,4};
        name = "nonempty name";
    }
    Board getcopy(){
        return *this;
    }
    int AttrLen(){
        return attribute.size();
    }
};

class Node{
    Board* board;

    Node* parent;

    std::vector<Node*> children;

    int depth=0;

    int score=0;

    bool terminal=false;
public:
    Node(Node* _parent, Board* _board,int _depth){
        parent = _parent;
        board = _board;

        depth = _depth;

        // randomize score
        score = rand() % 10;

        if(depth<2){
            for(int _=0;_<2;_++){
                Board new_state = board -> getcopy();
                children.push_back(new Node(this,&new_state,depth+1));
            }
        } else {
            children = {};
            terminal=true;
        }
    }
    int getScore(){
        return score;
    }
    bool isTerminal(){
        return terminal;
    }
    Node* getBestChild(){
        if(!terminal){
            if(children[0] ->getScore() > children[1] -> getScore()){
                return children[0];
            } else {
                return children[1];
            }
        } else {
            return nullptr;
        }
    }
    Board* getStateptr(){
        return board;
    }

};

int main(){

    // make a board
    Board my_board = Board();

    // make a root node
    Node root = Node(nullptr, &my_board, 0);

    Node* best_child = root.getBestChild();
    while(!best_child -> isTerminal()){
        best_child = best_child -> getBestChild();
    }

    cout << "Best leaf by tree traversal has score: " << best_child -> getScore() << endl;
    cout << "Best leaf associated state has " << best_child -> getStateptr() ->AttrLen() << " values in its attribute vector though! (Should be 4)" << endl;

}
    

我正在实现一种树搜索，该树搜索需要能够从树中获取“最有希望的节点”，然后对该节点执行某些操作，以便为...更新树的其余部分...]]] >

不确定这是否是唯一的问题，但在这里

for(int _=0;_<2;_++){
    Board new_state = board -> getcopy();
    children.push_back(new Node(this,&new_state,depth+1));
}
您创建的Board副本仅存在于for循环中。它会在}自动销毁。因此，您存储在节点中的指针是悬空的。指向已久的对象。

请注意，您的getcopy有点奇怪。您应该改为使用复制构造函数。通常，当您的类型通过原始指针管理资源时（不确定是否确实如此），则需要遵守3/5规则。在任何情况下，您都可以通过编写Board new_state = *board;来简单地复制电路板（假设编译器生成的复制构造函数执行正确的操作）。

Answer 1

不确定这是否是唯一的问题，但在这里