我正在制作一个游戏，球从智能手机屏幕的每一侧随机产生，尽管球主要在屏幕的顶部产生，在屏幕的底部和侧面有 2/10 次产生，但我得到了它的工作。代码如下:
//creates the balls and sets their position as well as the random timer for each
private void spawnBalls1() {
Rectangle ball1 = new Rectangle();
ball1.x = MathUtils.random(639, 641);
ball1.y = 720;
ball1.width = 32;
ball1.height = 32;
balls1.add(ball1);
lastDropTime = TimeUtils.nanoTime();
}
private void spawnBalls2() {
Rectangle ball2 = new Rectangle();
ball2.x = 0;
ball2.y = MathUtils.random(359, 361);
ball2.width = 32;
ball2.height = 32;
balls2.add(ball2);
lastDropTime = TimeUtils.nanoTime();
}
private void spawnBalls3() {
Rectangle ball3 = new Rectangle();
ball3.x = MathUtils.random(639, 641);
ball3.y = 0;
ball3.width = 32;
ball3.height = 32;
balls3.add(ball3);
lastDropTime = TimeUtils.nanoTime();
}
private void spawnBalls4() {
Rectangle ball4 = new Rectangle();
ball4.x = 1280;
ball4.y = MathUtils.random(359, 361);
ball4.width = 32;
ball4.height = 32;
balls4.add(ball4);
lastDropTime = TimeUtils.nanoTime();
}
//draws the balls
for (Rectangle ball1 : balls1) {
batch.draw(Ball, ball1.x, ball1.y);
}
for (Rectangle ball2 : balls2) {
batch.draw(Ball, ball2.x, ball2.y);
}
for (Rectangle ball3 : balls3) {
batch.draw(Ball, ball3.x, ball3.y);
}
for (Rectangle ball4 : balls4) {
batch.draw(Ball, ball4.x, ball4.y);
}
// if the time minus the time of the last ball spawn is less than x then spawn another ball in a random place
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls1();
Iterator<Rectangle> iter1 = balls1.iterator();
while(iter1.hasNext()) {
Rectangle balls1 = iter1.next();
balls1.y -= 500 * Gdx.graphics.getDeltaTime();
if (balls1.overlaps(square)) {
score++;
showScore = "Score: " + score;
iter1.remove();
}
}
// if the time minus the time of the last ball spawn is less than x then spawn another ball in a random place
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls2();
Iterator<Rectangle> iter2 = balls2.iterator();
while (iter2.hasNext()) {
Rectangle balls2 = iter2.next();
balls2.x += 500 * Gdx.graphics.getDeltaTime();
if (balls2.overlaps(square)) {
score++;
showScore = "Score: " + score;
iter2.remove();
}
}
// if the time minus the time of the last ball spawn is less than x then spawn another ball in a random place
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls3();
Iterator<Rectangle> iter3 = balls3.iterator();
while(iter3.hasNext()) {
Rectangle balls3 = iter3.next();
balls3.y += 500 * Gdx.graphics.getDeltaTime();
if (balls3.overlaps(square)) {
score++;
showScore = "Score: " + score;
iter3.remove();
}
}
// if the time minus the time of the last ball spawn is less than x then spawn another ball in a random place
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls4();
Iterator<Rectangle> iter4 = balls4.iterator();
while(iter4.hasNext()) {
Rectangle balls4 = iter4.next();
balls4.x -= 500 * Gdx.graphics.getDeltaTime();
if (balls4.overlaps(square)) {
score++;
showScore = "Score: " + score;
iter4.remove();
}
}
我应该做什么/改变才能让它们在每一侧随机生成？
提前致谢!
最佳答案
所有四种球生成方法共享相同的 lastDropTime 变量。执行时间
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls1();
和
if (TimeUtils.nanoTime() - lastDropTime > 1000000000) spawnBalls4();
将是 1000000000 ns 的一小部分，因此 spawnBalls2-4 不太可能被调用。
您应该只进行一项测试而不是四项，并随机选择一种方法。像这样的事情:
if (TimeUtils.nanoTime() - lastDropTime > 1000000000){
switch (MathUtils.randomInt(4)){
case 0: spawnBalls1(); break;
case 1: spawnBalls2(); break;
case 2: spawnBalls3(); break;
case 3: spawnBalls4(); break;
}
}
此外，这是大量的复制粘贴代码，这会增加您的时间投入。您应该编写一个 spawnBalls 方法，该方法采用一些影响球放置位置的参数。
关于java - 随机产生球，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35276354/