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编程康威的生活游戏

[英]Programming Conway's Game of Life

so im programming conways game of life in GUI form. 因此,即时通讯程序以GUI形式简化了生活游戏。 the output is not correct and i dont know why. 输出不正确,我不知道为什么。 this is the code that handles the "next generation". 这是处理“下一代”的代码。

for(int i=0; i < ROW; i++) {
    for(int j=0; j < COL; j++) {
        if(i > 0 && i < ROW-1 && j > 0 && j < COL -1) {
            if(grid.getButton(i-1, j-1).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i-1, j).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i-1, j+1).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i, j-1).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i, j+1).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i+1, j-1).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i+1, j).getBackground() == Color.BLUE) liveNeighbor++;
            if(grid.getButton(i+1, j+1).getBackground() == Color.BLUE) liveNeighbor++;

            if(grid.getButton(i, j).getBackground() == Color.WHITE) {
                if(liveNeighbor == 3) 
                    newGrid.getButton(i, j).setBackground(Color.BLUE);
                } else {
                    if(liveNeighbor > 3 || liveNeighbor < 2) 
                        newGrid.getButton(i, j).setBackground(Color.WHITE);
            }
            liveNeighbor=0;
        }
    }
}

maybe im missing something, but im pretty sure this is right. 也许我错过了一些东西,但是我很确定这是正确的。 any suggestions? 有什么建议么?

This is the problem: 这就是问题:

if(grid.getButton(i, j).getBackground() == Color.WHITE) {
    if(liveNeighbor == 3) 
        newGrid.getButton(i, j).setBackground(Color.BLUE);
    } else {
        if(liveNeighbor > 3 || liveNeighbor < 2) 
            newGrid.getButton(i, j).setBackground(Color.WHITE);
}

Your code only sets the color if the current tile is dead. 您的代码在当前图块已死的情况设置颜色。 If it's alive, it will never become dead. 如果它还活着,它将永远不会死。 I would write this as: 我可以这样写:

boolean currentlyAlive = grid.getButton(i, j).getBackground() == Color.BLUE;
boolean aliveNextStep = (liveNeighbour == 3) ||
                        (currentlyAlive && liveNeighbor == 2);
newGrid.getButton(i, j).setBackground(aliveNextStep ? Color.BLUE : Color.WHITE);

I'd also separate the logical data (a grid of boolean values) from the display part - the "business logic" here shouldn't know anything about colors. 我还要从显示部分分离逻辑数据(布尔值的网格)-这里的“业务逻辑”应该对颜色一无所知。

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