卡在JS中编程Conway的"Game of Life"

Question

我们必须为学校项目编写一个JavaScript版本的Conway's Game of Life，但是我们被困在了循环边上。整个过程运行良好，但是计算邻居数量的函数对边缘上的单元格不起作用(因为它必须计算数组外的值，这些值是未定义的)。我们尝试了几个选项，但它们都改变了程序其余部分的功能。

我们应该添加什么使它在网格边缘工作?

    var totalNeighbors = function(x, y) {
    var total = 0;
    if (x > 0 && cells[(x - 1)][y] == 1) {
        total++;
    }
    if (x < (width - 1) && cells[x + 1][y] == 1) {
        total++;
    }
    if (y > 0 && cells[x][y - 1] == 1) {
        total++;
    }
    if (y < (height - 1) && cells[x][y + 1] == 1) {
        total++;
    }
    if (y > 0 && x > 0 && cells[x - 1][y - 1] == 1) {
        total++;
    }
    if (y > 0 && x < (width - 1) && cells[x + 1][y - 1] == 1) {
        total++;
    }
    if (y < (height - 1) && x > 0 && cells[x - 1][y + 1] == 1) {
        total++;
    }
    if (y < (height - 1) && x < (width - 1) && cells[x + 1][y + 1] == 1) {
        total++;
    }
    return total;
};

谢谢!

Answer 1

我会这样写:
_{如你所见，我重构了一点}

var isvalid = function(x, y) {
        /*
         * This returns 1 if cells[x][y] == 1.
         * Otherwise, we return 0.
         * NOTE: If cells[x, y] is out of bounds, we return 0.
         * GLOBALS USED: cells, width, and height.
         */
        //This returns true if (index < size && index >= 0)
        //Used to check that index is not an invalid index.
        var inbounds = function (size, index) {
                return (index >= 0 && index < size);
        };
        //given point is out of bounds
        if (!inbounds(width, x) || !inbounds(height, y)) {
                return 0;
        }
        //everything is good
        return (cells[x][y] === 1) ? 1 : 0;
    };
var totalNeighbors = function(x, y) {
    var total = 0;
    //cells[x-1][y]
    total += isvalid(x-1, y);
    //cells[x + 1][y]
    total += isvalid(x+1, y);
    //cells[x][y - 1]
    total += isvalid(x, y-1);
    //cells[x][y + 1]
    total += isvalid(x, y+1);
    //cells[x - 1][y - 1]
    total += isvalid(x-1, y-1);
    //cells[x + 1][y - 1]
    total += isvalid(x+1, y-1);
    //cells[x - 1][y + 1]
    total += isvalid(x-1, y+1);
    //cells[x + 1][y + 1]
    total += isvalid(x+1, y+1);
    return total;
};

---

PS:您的原始代码示例是37行，没有注释。我的代码示例是52行带有注释的和33行不带有注释的。

据我所知，这种方法更简洁，更短。div;)