Ios 当UIScrollView子视图与UIAPTgestureRecognitor一起使用时，它会释放（删除）内存

该示例非常简单，一组UIImageView作为子视图添加到UIScrollView：

轻触UIImageView后，会将其发送到wideDisplay:方法以广泛显示：

-(void)wideDisplay:(UITapGestureRecognizer*)gesture{

    UAModalPanel *wideScreen = [[UAModalPanel alloc] initWithFrame:self.view.bounds];
    wideScreen.padding = UIEdgeInsetsMake(20.0, 20.0, 20.0, 20.0);

    self.imageView = [[UIImageView alloc]init];
    self.imageView = (UIImageView*)gesture.view;
    self.imageView.frame = CGRectMake(20, 20, 790, 500);

    [wideScreen.contentView addSubview:self.imageView];
    [self.view addSubview:wideScreen];    

}

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将imageView添加到模式视图后，此UIImageView将从scrollview子视图中删除。imageView是一个强大的属性，但尽管如此，在添加到自定义模式视图后，它似乎失去了引用，如何防止在选择它后将其从scrollview中删除？

每次调用-wideDisplay:？

是，如下所示，每次我选择UIImageView并调用wideDisplay:方法时，都会实例化self.imageView。我不明白为什么在使用ITView后会从scrollview中释放子视图一次只能有一个superview。您可以将图像视图从第一个代码段中添加的self.scrollView移动到widescreen.contentView。如果将其移动到widescreen.contentView，它将从self.scrollview中删除。此外，您还创建了一个视图[[UIImageView alloc]init]，然后通过重新指定self.imageView立即放弃该视图

-(void)wideDisplay:(UITapGestureRecognizer*)gesture{

    UAModalPanel *wideScreen = [[UAModalPanel alloc] initWithFrame:self.view.bounds];
    wideScreen.padding = UIEdgeInsetsMake(20.0, 20.0, 20.0, 20.0);

    self.imageView = [[UIImageView alloc]init];
    self.imageView = (UIImageView*)gesture.view;
    self.imageView.frame = CGRectMake(20, 20, 790, 500);

    [wideScreen.contentView addSubview:self.imageView];
    [self.view addSubview:wideScreen];    

}