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如何改变FloatBuffer的内容,保持业绩?
原标题:how to change the content of a FloatBuffer, keeping performance?

i. 利用审判

floatbuffer.put(float[]);

但是,由于我处理200多个广场,所有有逐个框架更新的精彩文字坐标,我的微薄会急剧下降,游戏变得太大不易变。

y 认为提到的关于坏学的方法,即使用浮水,使用无动于衷,但同样,在缓冲方法“投射”时,速度也比较缓慢。

因此,我如何能够以最佳业绩更新所有浮标者?

EDIT:即解决我的问题,“投射”方法本身并不缓慢,问题在于每台浮游艇开始投放新浮标,而只是改变浮标所含每个元素的价值,避免了许多GC活动。

问题回答

有一些改进业绩的方法。 简言之,这里有几个方面:

1) 启动浮标并实行每个框架的转变:——如果你的平方正越过屏幕,将其置于浮标线上,然后对每个框架分别实行一个矩阵转变。 在这一方法中,你一劳永逸地填平了浮雕,只是更新了每一方的转变。

<>2> 浮面中唯一独特的形状—— 如果你的任何广场相互重叠,则将每个框架称为同一个缓冲,以保存记忆。

<>strong>3 参考文献目录 它与你一样,可以这样做,但将数据储存在垂直的缓冲中,并利用阵列的指数把vert指。

如果你仍然有问题,我建议看一下这个理论:





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