Question

基本上，我有一个制作正方形的程序，并将左，右，顶部和底部存储在一个数组中。当它成为一个新的正方形时，它会在数组中循环。如果AABB碰撞检测使新正方形与另一个正方形重叠，则应确保不显示正方形，然后再次尝试。这是我制作的代码片段，我认为问题在于：

var xTopsBotsYTopsBotsSquares = [];

//Makes a randint() function.
function randint(min, max) {

    return Math.floor(Math.random() * (max - min + 1)) + min;
}

function checkForOccupiedAlready(left, top, right, bottom) {

    if (xTopsBotsYTopsBotsSquares.length == 0) {

        return true;
    }
    for (i in xTopsBotsYTopsBotsSquares) {
        if (i[0] <= right || i[1] <= bottom ||
            i[2] >= left || i[3] >= top) {/*Do nothing*/} 
        else {

            return false;
        }
    }

    return true;
}

//Makes a new square
function makeNewsquare() {
    var checkingIfRepeatCords = true;

    //DO loop that checks if there is a repeat.
    do {
        //Makes the square x/y positions 
        var squareRandomXPos = randint(50, canvas.width - 50);
        var squareRandomYPos = randint(50, canvas.height - 50);

        //Tests if that area is already occupied
        if (checkForOccupiedAlready(squareRandomXPos, 
                                    squareRandomYPos, 
                                    squareRandomXPos+50, 
                                    squareRandomYPos+50) == true) {

            xTopsBotsYTopsBotsSquares.push([squareRandomXPos, 
                                            squareRandomYPos, 
                                            squareRandomXPos+50, 
                                            squareRandomYPos+50]);

            checkingIfRepeatCords = false;

        }
    }
    while (checkingIfRepeatCords == true);

}

非常感谢任何帮助。

Answer 1

我认为你的循环不正确，因为你使用i作为一个值，而它是一个关键：

for (i in xTopsBotsYTopsBotsSquares) {
    if (i[0] <= right || i[1] <= bottom ||
        i[2] >= left || i[3] >= top) {/*Do nothing*/} 
    else {

        return false;
    }
}

可能成为：

for (var i = 0, l < xTopsBotsYTopsBotsSquares.length; i < l; i++) {
    var data = xTopsBotsYTopsBotsSquares[i];

    if (data[0] <= right || data[1] <= bottom ||
        data[2] >= left  || data[3] >= top) {/*Do nothing*/} 
    else {

        return false;
    }
}

Javascript画布，通过数组的碰撞检测不起作用

1 个答案: