Question

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。

我做错了什么？

由于

/ /抱歉，我的英语来自谷歌

Answer 1

检查recognizer是否不是nil。向nil发送消息通常会返回nil或0，具体取决于返回类型，但是当它是结构返回类型时，结果是未定义的，并且可能（除其他外）导致崩溃。

Answer 2

首先，您似乎错过了设置frame的{{1}}。

Answer 3

我认为MyImg变量仅在imagePickerController中声明，因此在getTouchColor中无法访问，您需要在调用getTouchColor时包含对该视图的引用，否则它不会认可MyImg。

来自UITapGestureRecognizer的CGPoint

3 个答案: