Question

所以我试图在this教程之后制作一个Tic Tac Toe游戏。当我运行它并查看Chrome开发工具时，它会显示Uncaught RangeError: Maximum call stack size exceeded，并指出我对此函数的行：

var State = function(oldState) {
    this.turn = "";
    this.oMovesCount = 0;
    this.result = "still running";
    this.board = [];


    //get the information from the previous state to use it for following states
    if (typeof oldState !== "undefined") {
        var len = oldState.board.length;
        this.board = new Array(len);
        for (var i = 0; i < len; i++) {
            this.board[i] = oldState.board[i];
        }
        this.oMovesCount = oldState.oMovesCount;
        this.result = oldState.result;
        this.turn = oldState.turn;
    }

    //change to X or O accordingly
    this.advanceTurn = function() {
        //Was it just X's turn? If so, change to O. If not, change to X.
        this.turn = this.turn === "x" ? "o" : "x";
    };

    //checks for victory
    this.result = "still running";
    this.isTerminal = function() {
        var B = this.board;

        //check to see if there has been a victory
        //check rows
        for(var i = 0; i <= 6; i = i + 3) {
            if(B[i] !== "E" && B[i] === B[i+1] && B[i+1] == B[i+2]) {
                this.result = B[i] + " has won!";
                return true;
            }
        }
        //check columns
        for(var i = 0; i <= 2 ; i++) {
            if(B[i] !== "E" && B[i] === B[i+3] && B[i+3] === B[i+6]) {
                this.result = B[i] + " has won!";
                return true;
            }
        }
        //check diagonals
        for(var i = 0, j = 4; i <= 2 ; i = i + 2, j = j - 2) {
            if(B[i] !== "E" && B[i] == B[i+j] && B[i+j] === B[i + 2*j]) {
                this.result = B[i] + " has won!";
                return true;
            }
        };

        //if there have been no wins, check the number of empty cells
        //if there are no empty cells, it's a draw
        var available = this.emptyCells();
        if (available.length == 0) {
            this.result = "draw";
            return true;
        }
        else {
            return false;
        }
    };
    //keeps track of how many empty cells there are on the board
    this.emptyCells = function() {
        var indxs = [];
        for (var i = 0; i < 9; i++) {
            if (this.board[i] === "E") {
                indxs.push(i);
            }
        }
        return indxs;
    } 
};

我不明白为什么。 Here's完整代码，当您在其中一个单元格上单击Play然后OK时，会显示错误。 Here如果有帮助，它会托管在其他网站上。

谢谢！

Answer 1

AIAction方法中有一个拼写错误：

this.oMovesPosition = pos; //the position on the board where the O would be placed
this.minimaxVal = 0; //the minimax value of the state that the action leads to

this.applyTo = function(state) {
    var next = new State(state);

    //if the current turn in the current state is O, increment .oMovesCount
    next.board[this.movePosition] = state.turn;
    if (state.turn === "o") {
        next.oMovesCount++;
    }
    next.advanceTurn();
    return next;
};

注意第一行中的this.oMovesPosition，但是applyTo方法改为引用this.movePosition。

Answer 2

太递归，在小提琴行395中，你在递归中调用函数minimax。

var nextScore = miniMax(nextState);

必须在内存不足之前停止递归，或将递归转换为循环。

Answer 3

在AIaction = function(pos)中，您有两个拼写相同的movePosition属性：

this.oMovesPosition = pos;

和

next.board[this.movePosition] = state.turn;

因为它们是不同的属性，所以第二行总是等同于：

next.board[undefined] = state.turn;

......所以董事会从未真正改变过。因此，董事会永远不会被视为终端，您的递归永远不会停止。

＆＃34;超出最大呼叫堆栈大小＆＃34;错误

3 个答案: