Iphone 来自UITapGestureRecognitor的CGPoint
我需要获取UITapGestureRecognitor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Iphone 来自UITapGestureRecognitor的CGPoint,iphone,objective-c,ios,Iphone,Objective C,Ios,我需要获取UITapGestureRecognitor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:Y
-(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];
}
我的活动也会变得很有趣
-(void)getTouchColor:(UITapGestureRecognizer *) recognizer
{
if (recognizer.state==UIGestureRecognizerStateEnded)
{
CGPoint point = [recognizer locationInView:MyImg];
NSLog(@"%@", NSStringFromCGPoint(point));
}
如果我拆下这条线
CGPoint point = [recognizer locationInView:MyImg];
代码工作正常,应用程序不会崩溃
我做错了什么
谢谢
/很抱歉,我的英文来自谷歌第一件事似乎是,您没有设置
MyImg的框架
。MyImg
请检查识别器
是否为nil
。将消息发送到nil
通常会返回nil
或0,具体取决于返回类型,但如果是结构返回类型,则结果未定义,并且可能(除其他外)导致崩溃。我认为MyImg变量仅在imagePickerController中声明,因此在getTouchColor
中无法访问,调用getTouchColor
时需要包含对该视图的引用,否则它将无法识别MyImg
Jeff如果线条快照Gpoint正确工作,它不会返回null感谢我更改了创建帧的代码,但应用程序继续关闭感谢