Question

我有类Node定义：

class Node
{
   public:
      Node(string newName);
      Node();
      void setNodeName(string newName);
      string getNodeName();
      void attachNewNode(Node *newNode, int direction);
      Node *getAttachedNode(int direction);
   private:
      string name;
      Node *attachedNodes[4];
};

Node::Node(string newName)
{
   name = newName;
}

Node::Node()
{};

void Node::setNodeName(string newName)
{
   name = newName;
}

string Node::getNodeName()
{
   return name;
}

void Node::attachNewNode(Node *newNode, int direction)
{
   attachedNodes[direction] = newNode;
}

Node* Node::getAttachedNode(int direction)
{
   return attachedNodes[direction];
}

我有一个文件Maze1.txt：

9
A1
C3
A1 A2 B1 * *
A2 * B2 A1 *
A3 * B3 * *
B1 * * * A1
B2 B3 C2 * A2
B3 * * B2 A3
C1 C2 * * *
C2 C3 * C1 B2
C3 * * C2 *

其中9是要创建的节点数，A1是我们将开始导航的节点，C3是我们将尝试查找路径的节点，以下行代表节点本身和它们具有的指针与他们相关联。例如：

A1 A2 B1 * *

表示节点A1的指针指向北方的节点A2，东方的B1，南方的空白，西方的空白。

A2 * B2 A1 *

表示节点A2的指针指向北方的节点null，东方的B2，南方的A1，以及西方的null。

我正在尝试创建一个“构建”节点“迷宫”的功能。以下内容将私有变量Nodes startNode和endNode设置为各自的节点，numNodes设置为文件给出的节点数。

如何处理字符串数据以为所有节点标题创建节点，然后在适当的位置指定指针。尝试：

ifstream instream;
instream.open("Maze1.txt");
string line;
string data;
int numLines = 1;
int numNodes;
Node startNode();
Node endNode();

while(getline(instream, line))
{
   istringstream iss(line);
   data += line + "\n";
   iss.clear();

   if(numLines == 1)
   {
      istringstream buffer(line);
      buffer >> numNodes;
   }
   if(numLines == 2)
      Node startNode(line);
   if(numLines == 3)
      Node endNode(line);

   if(numLines > 3)
   {
      Node temp(line.substr(0,2));
      rooms.push_back(temp);
   }

   iss.clear();
   numLines++;
}

这将创建并填充节点向量，每个节点都命名为文件的每个字符串行中提到的第一个节点。在这个循环之后，我需要运行另一个循环来查看每个字符串，并为向量中的相应节点指定指针。尝试：

ifstream repeat;
repeat.open(filename);
numLines = 1;
skipBlanks = 1;
int roomNum = 0;

while(getline(repeat, line))
{
   if(line.empty())
   {}
   else
   {
      istringstream iss(line);

      if(numLines == 1)
         skipBlanks++;
      if(numLines == 2)
         skipBlanks++;
      if(numLines == 3)
         skipBlanks++;

      if(numLines > 3 && skipBlanks > 3)
      {
         int first = line.find(" ", 0);
         int second = line.find(" ", first + 1);
         int third = line.find(" ", second + 1);
         int fourth = line.find(" ", third + 1);

         for(int i = 0; i < rooms.size(); i++)
         {
            if(rooms[i].getNodeName() == line.substr(first+1,2))
               rooms[roomNum].attachNewNode(rooms[i],1);
            if(rooms[i].getNodeName() == line.substr(second+1,2))
               rooms[roomNum].attachNewNode(rooms[i],2);
            if(rooms[i].getNodeName() == line.substr(third+1,2))
               rooms[roomNum].attachNewNode(rooms[i],3);
            if(rooms[i].getNodeName() == line.substr(fourth+1,2))
               rooms[roomNum].attachNewNode(rooms[i],4);
         }
      }

      roomNum++;
      numLines++;
      iss.clear();
   }
}

但是，每个attachNewNode（Node * newNode，int direction）函数调用都会出现编译错误。

error: no matching function for call to ‘Node::attachNewNode(__gnu_cxx::__alloc_traits<std::allocator<Node> >::value_type&, int)’

rooms[roomNum].attachNewNode(rooms[i],1);
                                       ^
note: candidate is:
note: void Node::attachNewNode(Node*, int) 
 void Node::attachNewNode(Node *newNode, int direction)
  ^
note:   no known conversion for argument 1 from '__gnu_cxx::__alloc_traits<std::allocator<Node> >::value_type {aka Node}’ to ‘Node*’

这是什么意思？我怎样才能更正指针的分配？

Answer 1

没有阅读完整的代码：错误说明Node::attachNewNode需要指向Node的指针，但您自己正在给他Node。通过从

更改您的电话可以解决此特定问题

     for(int i = 0; i < rooms.size(); i++)
     {
        if(rooms[i].getNodeName() == line.substr(first+1,2))
           rooms[roomNum].attachNewNode(rooms[i],1);
        if(rooms[i].getNodeName() == line.substr(second+1,2))
           rooms[roomNum].attachNewNode(rooms[i],2);
        if(rooms[i].getNodeName() == line.substr(third+1,2))
           rooms[roomNum].attachNewNode(rooms[i],3);
        if(rooms[i].getNodeName() == line.substr(fourth+1,2))
           rooms[roomNum].attachNewNode(rooms[i],4);
     }

到

     for(int i = 0; i < rooms.size(); i++)
     {
        if(rooms[i].getNodeName() == line.substr(first+1,2))
           rooms[roomNum].attachNewNode(&rooms[i],1);
        if(rooms[i].getNodeName() == line.substr(second+1,2))
           rooms[roomNum].attachNewNode(&rooms[i],2);
        if(rooms[i].getNodeName() == line.substr(third+1,2))
           rooms[roomNum].attachNewNode(&rooms[i],3);
        if(rooms[i].getNodeName() == line.substr(fourth+1,2))
           rooms[roomNum].attachNewNode(&rooms[i],4);
     }

即。您必须在四次调用中添加&，以便将指针传递给函数而不是对象本身。

从字符串文件c ++创建一个迷宫的节点

1 个答案: