Objective c 在Mac OS X下知道手指在轨迹板中的位置

我正在开发一个Mac应用程序，我想知道触摸板上手指的位置

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有可能吗？如果有，怎么做？

我不知道是否有ObjC接口，但您可能会发现C很有趣。

在Cocoa（Obj-C级别）尝试以下方法-尽管记住许多用户仍在使用鼠标控制

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您的视图需要设置为接受触摸（

[self-setAcceptsTouchEvents:YES]

）。当你得到一个触摸事件，如

-touchesBeganWithEvent:

，你可以通过查看手指的

标准化位置（范围是[0.0,1.0]x[0.0,1.0]），根据其
设备的大小来判断手指的位置（每英寸有72 bp）。轨迹板的左下角被视为零原点

例如：

- (id)initWithFrame:(NSRect)frameRect {
   self = [super initWithFrame:frameRect];
   if (!self) return nil;

   /* You need to set this to receive any touch event messages. */
   [self setAcceptsTouchEvents:YES];

   /* You only need to set this if you actually want resting touches.
    * If you don't, a touch will "end" when it starts resting and
    * "begin" again if it starts moving again. */
   [self setWantsRestingTouches:YES]
   return self;
}

/* One of many touch event handling methods. */
- (void)touchesBeganWithEvent:(NSEvent *)ev {
   NSSet *touches = [ev touchesMatchingPhase:NSTouchPhaseBegan inView:self];

   for (NSTouch *touch in touches) {
      /* Once you have a touch, getting the position is dead simple. */
      NSPoint fraction = touch.normalizedPosition;
      NSSize whole = touch.deviceSize;
      NSPoint wholeInches = {whole.width / 72.0, whole.height / 72.0};
      NSPoint pos = wholeInches;
      pos.x *= fraction.x;
      pos.y *= fraction.y;
      NSLog(@"%s: Finger is touching %g inches right and %g inches up "
            @"from lower left corner of trackpad.", __func__, pos.x, pos.y);
   }
}

（将此代码视为一个示例，而不是久经考验的真实示例代码；我只是将其直接写到注释框中。）
Swift 3:

我已经为NSTouch编写了一个扩展，它返回相对于NSView的触摸板触摸位置：

extension NSTouch {
    /**
     * Returns the relative position of the touch to the view
     * NOTE: the normalizedTouch is the relative location on the trackpad. values range from 0-1. And are y-flipped
     * TODO: debug if the touch area is working with a rect with a green stroke
     */
    func pos(_ view:NSView) -> CGPoint{
        let w = view.frame.size.width
        let h = view.frame.size.height
        let touchPos:CGPoint = CGPoint(self.normalizedPosition.x,1 + (self.normalizedPosition.y * -1))/*flip the touch coordinates*/
        let deviceSize:CGSize = self.deviceSize
        let deviceRatio:CGFloat = deviceSize.width/deviceSize.height/*find the ratio of the device*/
        let viewRatio:CGFloat = w/h
        var touchArea:CGSize = CGSize(w,h)
        /*Uniform-shrink the device to the view frame*/
        if(deviceRatio > viewRatio){/*device is wider than view*/
            touchArea.height = h/viewRatio
            touchArea.width = w
        }else if(deviceRatio < viewRatio){/*view is wider than device*/
            touchArea.height = h
            touchArea.width = w/deviceRatio
        }/*else ratios are the same*/
        let touchAreaPos:CGPoint = CGPoint((w - touchArea.width)/2,(h - touchArea.height)/2)/*we center the touchArea to the View*/
        return CGPoint(touchPos.x * touchArea.width,touchPos.y * touchArea.height) + touchAreaPos
    }
}

扩展NSTouch{
/**
*返回触摸到视图的相对位置
*注意：normalizedTouch是轨迹板上的相对位置。值的范围为0-1。并且是y翻转的
*TODO:如果触摸区域使用绿色笔划的矩形，则进行调试
*/
func pos（u-view:NSView）->CGPoint{
设w=视图.框架.尺寸.宽度
设h=view.frame.size.height
让touchPos:CGPoint=CGPoint（self.normalizedPosition.x，1+（self.normalizedPosition.y*-1））/*翻转触摸坐标*/
让deviceSize:CGSize=self.deviceSize
让deviceRatio:CGFloat=deviceSize.width/deviceSize.height/*找到设备的比率*/
让视图比率：CGFloat=w/h
变量接触面积：CGSize=CGSize（w，h）
/*均匀地将设备收缩到视图框*/
如果（deviceRatio>viewRatio）{/*设备比视图宽*/
touchArea.height=h/viewRatio
touchArea.width=w
}否则，如果（deviceRatio

这是我写的一篇关于macOS中的GestureHUD类的文章。还有一个现成扩展的链接：

例子：
这似乎是一个非常有趣的话题。我需要看看，迪珀。谢谢