'list iterator not dereferencable'

Question

   #include <iostream> 
using namespace std; 

#include <list> 


//A queue for the working set 
//x,y co-ords of the square, path length so far 
struct square {
    int x;
    int y; 
    int path_length;
};

list<square> workingset; 


//A 2D array of ints to represent the board (duplicates list)
int board[10][10]; 

void generatelegalmove(square node, int x_offset, int y_offset); 
void printboard(); 

void main()
{
    //Initialises the board 
    int i, j;
    for (i=0; i<10; i++)
    {
        for (j=0; j<10; j++)
        {
            board[i][j] = 0;
        }
    }

    //The goal position - a number we will never reach 
    board[8][8] = 1337; 

    bool goal_found = false; 

    //Sets up initial position 
    square temp = {3, 7, 1}; 

    //Put initial position in working set and duplicates list
    workingset.push_back(temp); 
    board[3][7] = 1; 

    //Loop (until a goal is found)
    while(!goal_found)
    {
        //Get the head node from the working set
        square nodetocheck = workingset.front();

        //Exit if the goal has been found
        if(board[nodetocheck.x][nodetocheck.y] == 1337)
        {
            goal_found = true;
            break; 
        }

        //Generate the legal moves 
        generatelegalmove(nodetocheck, -1, 0); //One square to the left
        generatelegalmove(nodetocheck, 0, -1); //One square up
        generatelegalmove(nodetocheck, 1, 0); //One square to the right
        generatelegalmove(nodetocheck, 0, 1); //One square down


        if(!workingset.empty())
        {
                //workingset.pop_front();
        }

    //End Loop
    }

    //Print the Board 
    printboard();
    while(true);

    //Trace back and print Trace back (once implemented) 

    //Print other info 

}


void generatelegalmove(square node, int x_offset, int y_offset)
{
    node.x = node.x + x_offset;
    node.y = node.y + y_offset; 
    node.path_length = node.path_length+1; 
    //Is this square on the board
    if((node.x >= 0) && 
       (node.x < 10) && 
       (node.y >= 0) && 
       (node.y < 10) && 
       //Is this square empty
       (board[node.x][node.y] == 0))
    {

        workingset.push_back(node);
        board[node.x][node.y] = node.path_length; 
        //Add to working set
        //Add to duplicates list 
    }
    //(If a graphical animation is added, do it here, by printing the new board after each one, then sleeping for a few seconds) 
} 

我得到运行时错误'list iterator not dereferencable'。

我假设这与在while循环中调用workingset.pop_front()有关，但我不确定应该采取什么措施来解决这个问题。

每个循环，我想从列表的前面获取节点，稍微使用它，然后从列表中删除该节点。

这是generatelegalmove（）的代码 - 正如你所看到的，如果新的方块在板上（即在数组的两个维度中在0-9范围内，并且方块为空），它将添加这个新节点到工作集和board [] []（实际上是一个有效的重复列表）

Answer 1

鉴于您提供的样本，我可以看到一件事是错的。在每次循环迭代结束时，弹出前节点。但是，如果goal_found为真，则只退出循环，表示行：

square nodetocheck = workingset.front();

...很可能访问一个空的工作集。显然你在这里调用了可能添加节点的其他函数，但是如果没有节点，这可能是个问题。

编辑：我认为您的代码没有添加其他节点，因为您使用的是按位和&运算符而不是逻辑和&&运算符，导致您的工作集不会获取节点。