Question

我的Flash游戏有点问题。每当他们到达时我的一系列鸟类（障碍物）让我们说x-800他们每次在阵列中的随机位置重生在起始位置并且效果很好。但每次循环时，鸟一对一地堆叠在一个阵列的第一个位置。这很奇怪。

 public function setUpBirds() {
            for (var i:int = 0 ;i< 10; i++) {
            var mcClip:Bird = new Bird();
            var yVal:Number = (Math.ceil(Math.random()*100));
            birds.push(mcClip);
            birds[i].x = 100 * i;
            birds[i].y = yVal * i;

            birdsContainer.addChild(mcClip);

        }
        }

private function moveBirds(event:Event):void {
            birdsContainer.x = birdsContainer.x -10;

            if (birdsContainer.x == -500) {
                birdsContainer.x = 500;

                setUpBirds();

                    }    

                }

有什么想法吗？

Answer 1

每当birdsContainer x 500 setUpBirds()时，您拨打setUpBirds，所以让我们一步一步看看发生了什么：（解释为注入代码的代码注释）

第一次for (var i:int = 0 ;i< 10; i++) { //a new bird is created 10 times var mcClip:Bird = new Bird(); var yVal:Number = (Math.ceil(Math.random()*100)); //you add it to the array birds.push(mcClip); //birds[1] properly refers to the item you just pushed into the array birds[i].x = 100 * i; birds[i].y = yVal * i; birdsContainer.addChild(mcClip); }运行：

birds

第一次，一切都很棒，你的for (var i:int = 0 ;i< 10; i++) { //create 10 more new birds (in addition to the last ones) var mcClip:Bird = new Bird(); var yVal:Number = (Math.ceil(Math.random()*100)); //add to the array (which already has 10 items in it) birds.push(mcClip); //so if i is 5, the item you just pushed is at birds[14] //birds[i] will refer to a bird you created the first time through //eg bird[0] - bird[9] depending on `i`, but really you want bird[10] = bird[19] this time around birds[i].x = 100 * i; //your moving the wrong bird birds[i].y = yVal * i; //the new birds you create, will have an x/y of 0 //since birds[i] doesn't refer to these new birds birdsContainer.addChild(mcClip); }数组现在有10个项目。

现在，第二次运行该函数：

birds

现在你看到了问题？您的mcClip数组现在有20个项目，因此您现在引用数组中的错误项目。

要解决此问题，只需在birds[birds.length-1].x = 100 * i var而不是数组上设置x / y，或者for (var i:int = 0 ;i< 10; i++) { //check if there is NOT a bird already at this position in the array if(birds.length <= i || !birds[i]){ //no bird yet, so create it and add it and push it var mcClip:Bird = new Bird(); birds.push(mcClip); birdsContainer.addChild(mcClip); } //now set the position of the bird var yVal:Number = (Math.ceil(Math.random()*100)); birds[i].x = 100 * i; birds[i].y = yVal * i; }使用添加到数组中的最后一项。

另一方面，你的表现会非常糟糕，很快就会创造10只新鸽。如果你不断创造新的鸟类，你需要摆脱那些老鸟。

似乎你可能想要做的只是在每个循环中重新定位现有的鸟类，所以看起来像这样：

这样，你只创造了10只鸟，而你只是在每个循环中重置那些鸟的{{1}}位置。

数组元素保持堆叠在数组的第一个位置

1 个答案: