function coordinate(x, y) {
this.x = x,
this.y = y
}
function generateBaracade(numBlocks) {
var coordinate = getRandomBaracadeCoordinate();
var xStyle = expandDirection.down;
var yStyle = expandDirection.right;
var xNumBlocks = findRandomFactor(numBlocks);
var yNumBlocks = numBlocks / xNumBlocks;
//figure out which way to expand the blocks
if (coordinate.x + xNumBlocks > xBlocks) {
xStyle = expandDirection.left;
} else {
xStyle = expandDirection.right;
}
if (coordinate.y + yNumBlocks > yBlocks) {
yStyle = expandDirection.down;
} else {
yStyle = expandDirection.up;
}
for (var i = 0; i <= xNumBlocks - 1; i++) {
for (var j = 0; j <= yNumBlocks - 1; j++) {
var tempBlock = Object.create(block);
tempBlock.type = "obstruction";
tempBlock.color = "grey";
tempBlock.illegalTerrain = true;
tempBlock.coordinate = new coordinate(coordinate.x + (i * xStyle), coordinate.y + (j * yStyle));
blockContainer[coordinate.x + (i * xStyle)][coordinate.y + (j * yStyle)] = tempBlock;
};
};
}
我在行上收到“未捕获的TypeError:对象不是函数”:
tempBlock.coordinate = new coordinate(coordinate.x + (i * xStyle), coordinate.y + (j * yStyle));
这很奇怪,因为我遵循mozilla指南进行操作:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/new
编辑:getRandomBaracadeCoordinate的源。 return语句正是我想要做的,并且执行时没有错误。
function getRandomBaracadeCoordinate() {
var x = Math.floor((Math.random() * (xBlocks)));
var y = Math.floor((Math.random() * (yBlocks)));
return new coordinate(x, y);
}
最佳答案
您通过在coordinate
的第一行上使用其他相同的名称来掩盖getBaracade
函数:
var coordinate = getRandomBaracadeCoordinate();
getRandomBaracadeCoordinate()
返回的内容都不是函数,因此new coordinate
会引发错误。