Is there any way at all that I can tell how hard the screen is being pressed

问题

I want to find the pressure of the touch. I just don\'t know how to accomplish that result with out jailbreaking it and getting the raw touch data. Does anyone know How I would do this?

回答1:

You cannot get the pressure from the SDK nor undocumented methods. However you can detect the size of touch with undocumented methods.

In the GSEvent, which is a lower-level representation of UIEvent, there is a structure known as GSPathInfo with members:

typedef struct GSPathInfo {
    unsigned char pathIndex;        // 0x0 = 0x5C
    unsigned char pathIdentity;     // 0x1 = 0x5D
    unsigned char pathProximity;    // 0x2 = 0x5E
    CGFloat pathPressure;               // 0x4 = 0x60
    CGFloat pathMajorRadius;        // 0x8 = 0x64
    CGPoint pathLocation;           // 0xC = 0x68
    GSWindowRef pathWindow;         // 0x14 = 0x70
} GSPathInfo;

We notice that there is a pathPressure and pathMajorRadius. I can assure you the pressure member is useless – it always gives 0. However pathMajorRadius does contain meaningful information – it gives the major radius of the touch in millimeters. You can therefore give an extremely rough estimation if it's a heavy touch or light touch from the radius.

  -(void)touchesBegan:(NSSet*)touches withEvent:(UIEvent*)event {
     GSEventRef gsevent = [event _gsEvent];
     GSPathInfo first_touch = GSEventGetPathInfoAtIndex(gsevent, 0);
     if (first_touch.pathMajorRadius >= 9)
        NSLog(@"huge (heavy) touch");
     else
        NSLog(@"little (light) touch");
  }

Let me warn you again this is undocumented and you should not use it in AppStore apps.

Edit: On 3.2 and above the pathMajorRadius of a GSPathInfo is also available as an undocumented property in UITouch:

@property(assign, nonatomic, setter=_setPathMajorRadius:) CGFloat _pathMajorRadius;

so the code above could be rewritten using pure Objective-C:

  -(void)touchesBegan:(NSSet*)touches withEvent:(UIEvent*)event {
     UITouch* any_touch = [touches anyObject];
     if (any_touch._pathMajorRadius >= 9)
        NSLog(@"huge (heavy) touch");
     else
        NSLog(@"little (light) touch");
  }

回答2:

As of iOS 8.0, UITouch has a public majorRadius property which tells you the approximate size of the touch.

回答3:

In iOS 3.2 and 4.0 you can get the value more directly like this:

UITouch* touch = ...// get the touch object
float radius = [[touch valueForKey:@"pathMajorRadius"] floatValue];

Still not App Store approved, but handy for custom stuff.

回答4:

It's not possible with the SDK. It does not expose that data in anyway. And the screen does not sense pressure, and wether it even reports touch "size" to the OS is unknown. So sadly, what you are trying to do is not possible on a legit application.

I know because I asked the same thing: Can the iPhone detect the size of a touch?

来源：https://stackoverflow.com/questions/2103447/is-there-any-way-at-all-that-i-can-tell-how-hard-the-screen-is-being-pressed

标签

iphone

uitouch

pressure