为什么我的对象没有通过segue正确传递？

Question

说我有2个控制器,BarViewController和FooViewController.

FooViewController有一个名为imageView的UIImageView的插座:

@property (nonatomic, weak) UIImageView *imageView;

BarViewController有一个UIButton按钮的插座.BarViewController有一个从这个按钮到FooViewController的segue,名为BarToFooSegue(在故事板中完成).

当我运行以下代码,并在FooViewController.imageView.image上调用NSLog时,结果为nil,我的图像将不会显示.为什么会这样？

// code in BarViewController
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{

    if([segue.identifier isEqualToString:@"BarToFooSegue"]){
        NSURL *photoUrl = @"http://www.randurl.com/someImage"; // assume valid url
        UIImage *image = [UIImage imageWithData:[NSData dataWithContentsOfURL:photoUrl]];
        UIImageView *imageView = [[UIImageView alloc] initWithImage:image];

        [segue.destinationViewController setImageView:imageView];
    }
}

我已经尝试将FooViewController.imageView设置为强而不是弱,但问题仍然存在:

@property (nonatomic, strong) UIImageView *imageView;

运行我的调试器,我注意到FooViewController中的imageView在prepareForSegue中正确更新:但随后在几行后重新更新为一些新分配的imageView,其中@property image设置为nil.我不确定控制流的哪一部分导致这种情况,因为它发生在用汇编语言编写的行中.

我通过向FooViewController添加UIImage属性来使我的代码工作:

@property (nonatomic, strong) UIImage *myImage;

并更改prepareForSegue:在BarViewController中传递图像而不是imageView:

// code in BarViewController
- (void)prepareForSegue:(UIStoryboardSegue *)segue sender:(id)sender{

    if([segue.identifier isEqualToString:@"BarToFooSegue"]){
        NSURL *photoUrl = @"http://www.randurl.com/someImage"; // assume valid url
        UIImage *image = [UIImage imageWithData:[NSData dataWithContentsOfURL:photoUrl]];

        [segue.destinationViewController setMyImage:image];
}

并修改viewWillAppear:在FooViewController中:

- (void)viewWillAppear:(BOOL)animated{
    [self.imageView setImage:self.myImage];
}

Answer 1

[segue.destinationViewController view];在设置图像之前调用,这将导致加载视图层次结构,然后设置您的插座.