小编Bru*_*ite的帖子

我需要获取UITapGestureRecognizer的x和y位置,但这样做会导致应用程序崩溃

这是我创建识别器的代码

-(void)imagePickerController:(UIImagePickerController *) picker didFinishPickingMediaWithInfo:(NSDictionary *) info
{

    [[picker parentViewController] dismissModalViewControllerAnimated:YES];

    UIImage * image =[info objectForKey:@"UIImagePickerControllerOriginalImage"];

    [image drawInRect:CGRectMake(0,0, 200, 400)];

    MyImg =[[UIImageView alloc] initWithImage:image]; 

    UITapGestureRecognizer *recognizer;

    MyImg.userInteractionEnabled=YES;

    recognizer = [[UITapGestureRecognizer alloc] initWithTarget:self action:@selector(getTouchColor:)];
    [MyImg addGestureRecognizer:recognizer];    
    [recognizer release];

    [self.view addSubview:MyImg];

    [picker release];

}

我的事件GetTouchColor

-(void)getTouchColor:(UITapGestureRecognizer *) recognizer
{
    if (recognizer.state==UIGestureRecognizerStateEnded)
    {

    CGPoint point = [recognizer locationInView:MyImg];

        NSLog(@"%@", NSStringFromCGPoint(point));
}

如果我删除该行

 CGPoint point = [recognizer locationInView:MyImg];

代码完美运行,应用程序不会崩溃.

我究竟做错了什么？

谢谢

/ /抱歉,谷歌的英语