tra*_*uan 0 multithreading objective-c grand-central-dispatch ios
我正在尝试使用GCD一个接一个地放置图像,如下所示
-(void)setUpImages {
NSArray *images = @[[UIImage imageNamed:@"blogger-icon.png"],
[UIImage imageNamed:@"gplus-icon.png"],
[UIImage imageNamed:@"facebok-icon.png"]
];
[images enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
dispatch_sync(dispatch_get_main_queue(), ^{
UIImageView *imageView = [[UIImageView alloc] initWithFrame:CGRectMake(80, idx * ((UIImage*)obj).size.height + idx*30 + 10, ((UIImage*)obj).size.width, ((UIImage*)obj).size.height)];
NSLog(@"index is %@",NSStringFromCGRect(imageView.frame));
[imageView setImage:(UIImage*)obj];
[self.view.layer addSublayer:imageView.layer];
sleep(1);
});
}];
}
Run Code Online (Sandbox Code Playgroud)
我正在使用,dispatch_sync因为我希望它会等到它的块完成(第一个图像放在屏幕上)然后第二个图像将是,所以第三个图像.现在所有的事情都发生在主线上.
但是,似乎我在中间遇到了僵局,我的逻辑在某些方面是错误的.
我需要帮助来了解这种情况.请帮忙.
来自以下文件dispatch_sync:
调用此函数并以当前队列为目标会导致死锁.
也就是说,我认为你有一个设计问题,你可以在不涉及GCD的情况下达到预期的效果.
由于您希望每隔k秒更新一次UI ,因此请避免使用sleep(1)延迟使用和调用方法performSelector:withObject:afterDelay:.
就像是
- (void)updateImageViewWithImageAtIndex:(NSNumber *)i {
UIImage * image = self.images[i.intValue];
UIImageView * imageView = [[UIImageView alloc] initWithFrame:CGRectMake(80, idx * image.size.height + idx*30 + 10, image.size.width, image.size.height)];
NSLog(@"index is %@",NSStringFromCGRect(imageView.frame));
[imageView setImage:image];
[self.view.layer addSublayer:imageView.layer];
if (i < images.count - 1) {
[self performSelector:@selector(updateImageViewWithImageAtIndex:) withObject:@(i.intValue++) afterDelay:1];
}
}
- (void)setUpImages {
// Assuming you have declared images as a property
self.images = @[[UIImage imageNamed:@"blogger-icon.png"],
[UIImage imageNamed:@"gplus-icon.png"],
[UIImage imageNamed:@"facebok-icon.png"]
];
[self updateImageViewFromImages:images index:0];
}
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
129 次 |
| 最近记录: |