UIActionSheet再次出现在iPad上后崩溃

Question

UIActionSheet在第二次出现后崩溃。

.H文件

..UIActionSheetDelegate>{

 UIActionSheet *popupQuery;
}
@property (nonatomic, retain) UIActionSheet *popupQuery;

.M文件

-(IBAction)showActionSheet:(id)sender {

if (popupQuery.visible) {
    NSLog(@"popupQuery isVisible");

    [popupQuery dismissWithClickedButtonIndex:-1 animated:YES];

    return;
}else{


popupQuery = [[UIActionSheet alloc] initWithTitle:@"Title" delegate:self cancelButtonTitle:@"Cancel" destructiveButtonTitle:@"Readibility" otherButtonTitles:@"Email URL", @"Print", nil];
popupQuery.actionSheetStyle = UIActionSheetStyleBlackOpaque;
//[popupQuery showInView:self.view];


[popupQuery showFromBarButtonItem:actionButton animated:YES];

[popupQuery release];
}
  }

Answer 1

在从actionButton显示popupQuery后释放它时，将放弃对该对象的所有权。如果对象被释放，那么popupQuery将指向一个被释放的对象，当你这样做的时候，popupQuery.visible可能会给你一个崩溃。既然你拥有它作为财产，你可以这样做-

self.popupQuery = [[[UIActionSheet alloc] initWithTitle:@"Title" delegate:self cancelButtonTitle:@"Cancel" destructiveButtonTitle:@"Readibility" otherButtonTitles:@"Email URL", @"Print", nil] autorelease];

删除末尾的release语句。现在，当您使用popupQuery.visible或self.poupQuery.visible时，对象将是有效的。在这种情况下，您将拥有所有权，并且可以安全地访问对象。

Answer 2

从UIActionSheet的documentation

- (void)dismissWithClickedButtonIndex:(NSInteger)buttonIndex animated:(BOOL)animated

参数

buttonIndex

单击的按钮的索引。按钮索引从0开始。

因此，问题可能出在您的-1。