Question

我有一个通过UIStoryboardPopoverSegue调用的viewcontroller，里面有一个调用imagePickerController的按钮，它可以用于第一次调用但第二次崩溃。

如果没有通过UIStoryboardPopoverSegue，代码就可以使用了。

   -(void)imagePickerController:(UIImagePickerController *)picker didFinishPickingMediaWithInfo:(NSDictionary *)info
{
    UIImage *image=[info objectForKey:UIImagePickerControllerOriginalImage];
    UIImageWriteToSavedPhotosAlbum (image, nil, nil , nil);
    [self dismissViewControllerAnimated:YES completion:nil];

}
-(IBAction)takephoto:(id)sender
{
    if ([UIImagePickerController isSourceTypeAvailable:UIImagePickerControllerSourceTypeCamera]) {

        self.imagePicker.delegate=self;
        self.imagePicker.sourceType=UIImagePickerControllerSourceTypeCamera;

        [self presentViewController:imagePicker animated:YES completion:nil];
    }

}

- (void)viewDidLoad
{
    [super viewDidLoad];

    if (!self.imagePicker)
    {
        self.imagePicker = [[UIImagePickerController alloc] init];
    }
}

我启用了NSZobmbie来显示任何可以帮助我跟踪它的日志。

* - [UIImagePickerController isKindOfClass：]：发送给的消息解除分配的实例0x1eb3b700

Answer 1

您的UIImagePickerController已发布，之后尝试访问 - 因此崩溃了！

要修复它，请使UIImagePickerController成为拥有实例的强引用属性：

@property (strong, nonatomic) UIImagePickerController *imagePicker;

初始化一次：

if (!self.imagePicker) self.imagePicker = [[UIImagePickerController alloc] init];

并使用：

[self.imagePicker doStuff]

进行访问。

那应该解决它。

修改

if ([[segue identifier]isEqualToString:@"tcwindshield"]) { self.ips = [segue destinationViewController]; self.ips.delegate = self self.ips.strStatValue=@"WindShield"; }

在UIStoryboardPopoverSegue中调用时，ImagePickerController崩溃presentViewController

1 个答案: