iPhone 开发 - 以编程方式创建的视图中的 viewDidLoad 和 viewDidUnload ?

发布于 2024-09-03 12:38:40 字数 225 浏览 3 评论 0原文

我在某处读到,在 UIViewController 中以编程方式创建的视图中,不使用 Interface Builder,不应使用 -viewDidLoad-viewDidUnload 。这是对的吗?为什么?我应该在哪里释放我保留属性的子视图?或者我不应该使用它们的属性?

编辑:阅读我对 Rob Napier 的回答的评论。

I read somewhere that in a programmatically created view in a UIViewController, not using Interface Builder, -viewDidLoad and -viewDidUnload should not be used. Is this right? Why? Where would I release subviews that I have retaining properties of? Or should I just not use properties for them?

EDIT: Read my comments on Rob Napier's answer.

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。

评论(2

久夏青 2024-09-10 12:38:40

-viewDidLoad 中创建子视图。如果您需要它们的 ivars,那么只需分配它们的值。通过将视图作为子视图添加到主视图来保留引用。

然后,当您的视图被卸载时,您应该将 ivars 设置为 nil,因为自从您的视图被删除并释放后,该对象已被释放。

因此,在您的标头

@interface MyViewController : UIViewController {
  IBOutlet UIView *someSubview; // assigned
}
@property (nonatomic, assign) IBOutlet UIView someSubview;
@end

和实现中,

@implementation MyViewController
//... some important stuff

- (void)viewDidLoad;
{
  [super viewDidLoad];
  someSubview = [[UIView alloc] initWithFrame:self.view.bounds];
  [self.view addSubview:someSubview]; // retains someSubview
  [someSubview release];   // we don't hold it
}

- (void)viewDidUnload;
{
  [super viewDidUnload];
  someSubview = nil;  // set the pointer to nil because someSubview has been released
}

//... more important stuff

@end

如果您希望也可以不在 -viewDidLoad 中释放 someSubview ,但您必须在 -viewDidUnload 中释放它code> 和 -dealloc 因为(如果我没记错的话)-viewDidUnload-dealloc 之前不会被调用。但如果您不保留 someSubview,则没有必要。

Create your subviews in -viewDidLoad. If you need ivars for them then only assign their values. The reference is hold by adding the views as subviews to you main view.

Then when your view is unloaded you should set your ivars to nil, because the object have been released since your view was removed and released.

So in your header

@interface MyViewController : UIViewController {
  IBOutlet UIView *someSubview; // assigned
}
@property (nonatomic, assign) IBOutlet UIView someSubview;
@end

And in your implementation

@implementation MyViewController
//... some important stuff

- (void)viewDidLoad;
{
  [super viewDidLoad];
  someSubview = [[UIView alloc] initWithFrame:self.view.bounds];
  [self.view addSubview:someSubview]; // retains someSubview
  [someSubview release];   // we don't hold it
}

- (void)viewDidUnload;
{
  [super viewDidUnload];
  someSubview = nil;  // set the pointer to nil because someSubview has been released
}

//... more important stuff

@end

If you wish you can also not release someSubview in -viewDidLoad, but then you have to release it in -viewDidUnload AND -dealloc since (if I remember right) -viewDidUnload isn't called before -dealloc. But this isn't necessary if you don't retain someSubview.

紧拥背影 2024-09-10 12:38:40

这里奇怪的是,未从 NIB 文件加载的 UIViewController 不会收到有关其视图卸载的通知(因此不会调用其 viewDidUnload 方法),除非您提供 loadView 方法的基本实现,例如:

- (void)loadView {
   self.view = [[[UIView alloc] initWithFrame:[UIScreen mainScreen].bounds] autorelease];
   [self.view setAutoresizingMask:UIViewAutoresizingFlexibleWidth | UIViewAutoresizingFlexibleHeight];
}

- (void)viewDidLoad {
   [super viewDidLoad];
   // create views...
}


- (void)viewDidUnload {
   // destroy views...
   [super viewDidUnload];
}

这仅发生在基本情况下UIViewController(例如 UITableViewController)不需要使用此解决方法进行修复。

所以罗布斯是对的。

the strange thing here is that an UIViewController not loaded from a NIB file is not notified about its view unloading (and so its viewDidUnload method is not called) unless you offer a base implementation of the loadView method, such as:

- (void)loadView {
   self.view = [[[UIView alloc] initWithFrame:[UIScreen mainScreen].bounds] autorelease];
   [self.view setAutoresizingMask:UIViewAutoresizingFlexibleWidth | UIViewAutoresizingFlexibleHeight];
}

- (void)viewDidLoad {
   [super viewDidLoad];
   // create views...
}


- (void)viewDidUnload {
   // destroy views...
   [super viewDidUnload];
}

this only happens to base UIViewController, an UITableViewController for example don't need to be fixed with this workaroud.

So Robs is right.

~没有更多了~
我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
原文