ios - 耗时操作前 View 不出现

Question

我已经在我的应用程序中实现了这个指标:https://github.com/vincechan/SwiftLoadingIndicator

在这个加载旋转器出现在前面之后，我希望它一直旋转，只要我的操作持续。

在我的操作中，我对图像做了一些处理，然后将它们显示给用户，但每次我单击“运行操作”按钮后，应用程序都会卡住几秒钟，用结果更新 View ，然后显示覆盖微调器.我试过这样的异步调度:

 @IBAction func manipulateImage(sender: AnyObject) {
    dispatch_async(dispatch_get_main_queue(), {
                    LoadingOverlayView.show()
                }); 
       if let beginImage = CIImage(image: self.imageView.image!) {
                    var outputImage = OutputImage(sourceImage: beginImage)
    //apply CIFilters:
                    outputImage.applyFilter(FilterType.Grayscale)
                    outputImage.applyFilter(FilterType.Sepia)
                    outputImage.applyFilter(FilterType.Vignette)
                    outputImage.applyFilter(FilterType.Shadow)

                    let cgimg = self.imageContext.createCGImage(outputImage, fromRect: outputImage.extent)

                    self.imageView.image = UIImage(CGImage: cgimg)
    LoadingOverlayView.hide()
}

但它不起作用。没有异步分派(dispatch)的普通方法调用的行为也完全相同。

Answer 1

您可能在 UI 线程上运行此代码。如果是这样，您可以按如下方式将操作分派(dispatch)到后台队列:

// we're already on the UI thread, so dispatch to a background queue
dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
   // do your background operations ...
   if let beginImage = CIImage(image: self.imageView.image!) {
       var outputImage = OutputImage(sourceImage: beginImage)
       //apply CIFilters:
       outputImage.applyFilter(FilterType.Grayscale)
       outputImage.applyFilter(FilterType.Sepia)
       outputImage.applyFilter(FilterType.Vignette)
       outputImage.applyFilter(FilterType.Shadow)

       let cgimg = self.imageContext.createCGImage(outputImage, fromRect: outputImage.extent)

       // make sure to dispatch UI function back to main queue!
       dispatch_async(dispatch_get_main_queue(), {
               self.imageView.image = UIImage(CGImage: cgimg)

                LoadingOverlayView.hide()
            }); 
     }
});

// we're already on the UI thread, so show the overlay now
LoadingOverlayView.show()