locationOfTouch和numberOfTouches

嗨我有这个识别器，设置2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 %fy %f",location.x, location.y); } 

据我所知，我用这两种方法循环触摸的数量，但我还没有想出如何：

 -(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]; 

touchIndex是0或1。