C# 堆栈推送调用问题
本文关键字：调用 问题 堆栈 | 更新日期: 2024-11-05 01:35:56
我遇到了一个问题，我的 C# 堆栈将接受推送的值，但随后也用新值覆盖堆栈中以前存在的元素。
下面是供参考的构造函数块：
public class MazNav //handles processing and navigation
{
private Maze m;
private static Stack<int[]> path;
private int[] currCell;
private int visCell;
private Random rng;
//constructor
public MazNav(Maze mz)
{
m = mz; //assigns this object to a maze
path = new Stack<int[]>(); //initialize the stack
currCell = m.getStart(); //starts the pathfinding at start cell
visCell = 1; //initializes the visited cells count
path.Push(currCell); //adds this cell to the stack
rng = new Random(); //initializes the randomizer
问题出现在此方法末尾的 if 块中（对不起，它仍然很丑;我正在调试中，一旦我有工作:)
public void buildMaze()
{
//variables to represent the current cell
int[] currPos; //coordinates
int nextDir = 0; //direction towards next cell
//variables to represent the next cell
int[] nextPos; //coordinates
int backDir = 0; //holds direction towards previous cell
bool deadEnd = false; //flags true when a backtrack is required
bool outOfBounds = false; //flags true when a move would leave the array
while (visCell < m.getTotCells()) //while the number of visited cells is less than the total cells
{
if (path.Count > 0) // if there is something in the stack
{
currPos = path.Peek(); //check the current coordinates
nextPos = currPos; //movement will happen one cell at a time; setting next cell coordinates the same as current allows easy adjustment
nextDir = findNextUnv(currPos); //find the direction of the next cell to check
deadEnd = false;
outOfBounds = false;
switch (nextDir)
{
case 0: //North
if (nextPos[0] - 1 >= 0)
{
nextPos[0]--;
backDir = 2;
}
else
{
outOfBounds = true;
}
break;
case 1: //East
if (nextPos[1] + 1 < m.getCols())
{
nextPos[1]++;
backDir = 3;
}
else
{
outOfBounds = true;
}
break;
case 2: //South
if(nextPos[0] + 1 < m.getRows())
{
nextPos[0]++;
backDir = 0;
}
else
{
outOfBounds = true;
}
break;
case 3: //West
if (nextPos[1] - 1 >= 0)
{
nextPos[1]--;
backDir = 1;
}
else
{
outOfBounds = true;
}
break;
case 99: //dead end
try
{
deadEnd = true;
path.Pop();
currPos = path.Peek();
int diff;
if (currPos[0] == nextPos[0])
{
diff = currPos[1] - nextPos[1];
if (diff == -1)
{
backDir = 3;
}
else if (diff == 1)
{
backDir = 1;
}
}
else if (currPos[1] == nextPos[1])
{
diff = currPos[0] - nextPos[0];
if (diff == -1)
{
backDir = 2;
}
else if (diff == 1)
{
backDir = 0;
}
}
m.getCell(nextPos[0], nextPos[1]).setBck(backDir, true);
}
catch (Exception) { }
break;
}
if (!deadEnd && !outOfBounds)
{
m.getCell(currPos[0], currPos[1]).setWal(nextDir, false);
m.getCell(nextPos[0], nextPos[1]).setWal(backDir, false);
path.Push(nextPos);
visCell++;
}
}
}
}e
推送调用只执行一次，但是当我在调试器上观察它时，在该行运行后，计数增加了 1，但堆栈中的每个元素现在都相同。 以前有人遇到过这种行为吗？ 我哪里出错了？
由于您对待nextPos和curPos的方式，您的代码中存在很多错误。
特别是这一行：
nextPos = currPos;
在这里，您将nextPos分配给与curPos相同的数组，因此如果您修改一个数组，另一个将随之修改。如果你将一个推送到堆栈上，然后修改任何一个，堆栈上的数组将随之修改。
我建议的解决方案是为您的位置使用不可变的数据类型，而不是数组（无论如何它都不是特别适合坐标）：
internal struct Point
{
internal int X { get; private set; }
internal int Y { get; private set; }
internal Point(int x, int y)
{
X = x;
Y = y;
}
internal Point NewX(int deltaX)
{
return new Point(X + deltaX, Y);
}
internal Point NewY(int deltaY)
{
return new Point(X, Y + deltaY);
}
}
这样，当您需要过渡到新点时，可以创建一个新点，而不是修改现有点：
if (nextPos.X - 1 >= 0)
{
nextPos = nextPos.NewX(-1);
backDir = 2;
}
这应该可以帮助您掌握您的值，并防止它们以各种方式相互覆盖。