尝试生成 n 个不碰撞的精灵； pygame 崩溃

Question

我对 PyGame 很陌生（两天前开始学习），我正在尝试制作一个游戏，在我的岛上生长随机数量的浆果。

我的（理想）算法会

1. 在静态浆果丛中生成
2. 检查它是否与我之前生成的任何浆果发生碰撞
3. 如果发生碰撞，请将其删除并生成另一个不发生碰撞的

我为 genBerries 函数编写了以下代码：

def genBerries(xRange, yRange, bushCount):  
    sucSpawns = 0
    berries = []
    while sucSpawns < bushCount:  # while you've spawned less than you want to
        spawnX = random.randint(xRange[0], xRange[1])  # randomise x coordinate for berry bush
        spawnY = random.randint(yRange[0], yRange[1])  # randomise y coordinate for berry bush
        berry = Berry(10, 5, 3, spawnX, spawnY)  # make a berry at this x and y
        for j in berries:
            collide = berry.rect.colliderect(j)  # check if berry colliding with some other berry
            if collide:
                berry.kill()  # if it is, get rid of it (and while loop will let u try again)
            else:
                sucSpawns += 1
                berries.append(berry)  # add this berry to the list of berries you've spawned successfully

xRange = island.rect.left, island.rect.right - 96  # defining x range berries can spawn in
yRange = (island.rect.top + 96, island.rect.bottom)  # defining y range berries can spawn in
bushCount = math.floor(random.normalvariate(conf['bA'], conf['bA']/5))  # defining how many berries spawn

genBerries(xRange, yRange, bushCount)

但是，一旦我这样做...

某种堆栈溢出（哈哈）似乎意味着这个进程没有终止？然后黑屏？？

如果我注释掉碰撞部分，它会给我（根据需要，但重叠）：

Answer 1

当您识别出碰撞时，您必须离开循环，并且只能将该对象插入到列表中一次，但不是针对每个不发生碰撞的对象：

while len(berries) < bushCount:
    spawnX = random.randint(xRange[0], xRange[1]) 
    spawnY = random.randint(yRange[0], yRange[1])
    berry = Berry(10, 5, 3, spawnX, spawnY)
    collide = None 
    for j in berries:
        collide = berry.rect.colliderect(j)
        if collide:
            break
            berry.kill()
    if collide = None:
        berries.append(berry)