objective-c - locationOfTouch 和 numberOfTouches

Question

嗨我有这个识别器，设置了 2 个触摸，但它只返回一个，而不是两个 CGPoint

-(void)gestureLoad {

UIGestureRecognizer *recognizer;

recognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(numTap2:)];
[(UITapGestureRecognizer *)recognizer setNumberOfTouchesRequired:2];
[self.view addGestureRecognizer:recognizer];
self.tapRecognizer = (UITapGestureRecognizer *)recognizer;
recognizer.delegate = self;
[recognizer release];

}

- (void)numTap2:(UITapGestureRecognizer *)recognizer {

CGPoint location = [recognizer locationInView:self.view];
NSLog(@"x %f y %f",location.x, location.y);

}

据我了解，我用这两种方法循环触摸次数，但我还没有弄清楚如何:

-(CGPoint)locationOfTouch:(NSUInteger)touchIndex inView:(UIView *)view {

}

-(NSUInteger)numberOfTouches {

}

非常感谢!

Answer 1

在 numTap2 中，使用:

CGPoint location = [recognizer locationOfTouch:touchIndex inView:self.view];

其中 touchIndex 为 0 或 1。