I m 在《行动》3中写上卡片游戏。 每张卡都有一门从闪电式4号出口电影展期的班子,其中包括卡片图形和单射。 当我用一张卡片时,我打电话给AndPlay,看看这部电影。
当框架速度放缓时,所有估计需要更长时间才能完成。 看来,通过不停的电影剪辑,可以确保纸上的所有框架。 因此,如果方案框架比率低于规定时间,则计算速度会放缓。
我总是以同样的速度 play升,因此总是在同一时间的屏幕上显示。 如果框架速度太慢,无法显示所有框架,则框架就会下降。 是否有可能以这种方式将闪闪电告诉囚犯? 如果不是,最容易地将这种行为本身列入方案?
最容易的方式是使用<代码> 日码(或setInterval,每1 000/fps milliseconds,其中注明MovieClip。
如果你需要更准确的话,你必须检查自上一次更新<>以来所花时间。 (stolen from the first comment of this article from Keith Peters blog, 其中解释了所有情况。)
这是自我使用闪电以来的一段时期,但从我理解的那段话来看,由于闪电使用了一种基于估算的模型。
然而,银星利用了基于时间的估算模式,并将像基准值所允许的那样,利用尽可能多的中间框架。
tw类(fl.transitions)。 Tween对这种情况是完美的。 通过将施工人的最后论点设定为TRUE,然后将其推论/触发点数与秒数挂钩,而且我知道它补偿了FPS的损失/加因。
您可以通过向建筑商传递一个新物体(Trigger(new Object()、“x”)来方便地将其与时段相提并论,并为工程部创建听众(TweenEvent)。 MoTION_FINISH,以激发你的活动!
我利用Tweens,做一件小事,因为事情容易、快捷,得到一切支持。 正如我说过的那样,它应当解决你的科索沃警察部队/滞后问题,因为它适应了科索沃警察部队的实时变化(我知道!)。
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