• iOS动画实现:弹簧效果


    源码地址:https://github.com/thermogl/TISpringLoadedViews,

    这个比较复杂,我写了个简化版的:

    https://www.dropbox.com/s/sv3yhm8dovh0adq/SpringDemo.zip

    - (void)simulateSpringWithDisplayLink:(CADisplayLink *)displayLink {
        
        if (springEnabled && !self.panning){
            CGPoint displacement = CGPointMake(self.center.x - restCenter.x,
                                               self.center.y - restCenter.y);
            //控件收到的力
            CGPoint kx = CGPointMake(springConstant * displacement.x, springConstant * displacement.y);
            //控件受到的阻力
            CGPoint bv = CGPointMake(dampingCoefficient    * velocity.x, dampingCoefficient * velocity.y);
            //加速度
            CGPoint acceleration = CGPointMake((kx.x + bv.x) / mass, (kx.y + bv.y) / mass);
            
            velocity.x -= (acceleration.x * displayLink.duration);
            velocity.y -= (acceleration.y * displayLink.duration);
            
            //设置控件新位置
            CGPoint newCenter = self.center;
            newCenter.x += (velocity.x * displayLink.duration);
            newCenter.y += (velocity.y * displayLink.duration);
            [self setCenter:newCenter];
        }
    }

    这个方法由CADisplayLink负责调用,每秒钟默认60次。要看懂这段代码,首先你要了解弹簧的原理,比如拉起弹簧收到的力怎么计算,往回弹收到的阻力怎么计算。

    下面是手势操作的代码,也就是拖动时触发的方法:

    - (void)viewWasPanned:(UIPanGestureRecognizer *)sender {
        
        CGPoint translation = CGPointApplyAffineTransform([sender translationInView:self.superview], CGAffineTransformMakeScale(panDragCoefficient, panDragCoefficient));
        CGPoint translatedCenter = CGPointMake(self.center.x + translation.x, self.center.y + translation.y);
        
        if (translation.x > 0 && (translatedCenter.x - restCenter.x) > panDistanceLimits.right){
            translation.x -= (translatedCenter.x - restCenter.x) - panDistanceLimits.right;
        }
        else if (translation.x < 0 && (restCenter.x - translatedCenter.x) > panDistanceLimits.left){
            translation.x += (restCenter.x - translatedCenter.x) - panDistanceLimits.left;
        }
        
        if (translation.y > 0 && (translatedCenter.y - restCenter.y) > panDistanceLimits.bottom){
            translation.y -= (translatedCenter.y - restCenter.y) - panDistanceLimits.bottom;
        }
        else if (translation.y < 0 && (restCenter.y - translatedCenter.y) > panDistanceLimits.top){
            translation.y += (restCenter.y - translatedCenter.y) - panDistanceLimits.top;
        }
        
        [self setCenter:CGPointMake(self.center.x + translation.x, self.center.y + translation.y)];
        [sender setTranslation:CGPointZero inView:self.superview];
    }
  • 相关阅读:
    toodifficult 题解
    Code Chef February Challenge 2019题解
    LOJ#3085. 「GXOI / GZOI2019」特技飞行(KDtree+坐标系变换)
    LOJ#3084. 「GXOI / GZOI2019」宝牌一大堆(递推)
    LOJ#3083. 「GXOI / GZOI2019」与或和(单调栈)
    LOJ#6046. 「雅礼集训 2017 Day8」爷(分块)
    LOJ#6045. 「雅礼集训 2017 Day8」价(最小割)
    LOJ#6044. 「雅礼集训 2017 Day8」共(Prufer序列)
    BZOJ4766: 文艺计算姬(Prufer序列)
    BZOJ3729: Gty的游戏(伪ETT)
  • 原文地址:https://www.cnblogs.com/shangdahao/p/3116731.html
Copyright © 2020-2023  润新知