尽快清除透明的 BufferedImage

发布于 2024-08-23 12:40:10 字数 639 浏览 16 评论 0原文

我有一个使用以下代码创建的透明 BufferedImage（我认为与它的创建方式无关）：

            GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
            GraphicsDevice gs = ge.getDefaultScreenDevice();
            GraphicsConfiguration gc = gs.getDefaultConfiguration();

            Rectangle screen = transformationContext.getScreen();

            // Create an image that supports transparent pixels
            return gc.createCompatibleImage((int) screen.getWidth(), (int) screen.getHeight(),
                    Transparency.BITMASK);

如何以最快的方式清除图像（与创建时状态相同的空图像）而不重新创建图像？重新创建映像会给 GC 带来负担，暂停 VM 并冻结 UI。

原文

I have a transparent BufferedImage created with the following code(not relevant how it is created, I think):

            GraphicsEnvironment ge = GraphicsEnvironment.getLocalGraphicsEnvironment();
            GraphicsDevice gs = ge.getDefaultScreenDevice();
            GraphicsConfiguration gc = gs.getDefaultConfiguration();

            Rectangle screen = transformationContext.getScreen();

            // Create an image that supports transparent pixels
            return gc.createCompatibleImage((int) screen.getWidth(), (int) screen.getHeight(),
                    Transparency.BITMASK);

How do I clear the image(empty image in the same state as it was created) in the fastest way possible without recreating the image? Recreating the image puts a burden on GC, pausing the VM and freezing the UI.

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紧拥背影 2024-08-30 12:40:10

明白了:)使用clearRect而不是用透明颜色填充。

            graphics = (Graphics2D) offlineBuffer.getGraphics();
            graphics.setBackground(new Color(255, 255, 255, 0));
            Rectangle screen = transformationContext.getScreen();
            graphics.clearRect(0,0, (int)screen.getWidth(), (int)screen.getHeight());

Got it :) used clearRect instead of fill with a transparent color.

            graphics = (Graphics2D) offlineBuffer.getGraphics();
            graphics.setBackground(new Color(255, 255, 255, 0));
            Rectangle screen = transformationContext.getScreen();
            graphics.clearRect(0,0, (int)screen.getWidth(), (int)screen.getHeight());

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顾冷 2024-08-30 12:40:10

一种相对较快的方法，但我不知道它是否是最快的（我想看到其他答案）是拥有另一张你从未修改过的图片，并且始终“完全清除”/“完全透明”，然后你做了一个光栅副本，假设你将该副本命名为“CLEAR”：

imageYouWantToClear.setData( CLEAR.getRaster() );

请注意，在性能方面，处理图形可能非常棘手，因为有很多没有很好记录的行为。例如，您的图像（例如 CLEAR 图像）可能是硬件加速的，但是一旦您使用变异方法（例如 setRgb( ））并且事实证明很难意识到您刚刚失去了硬件加速的好处。

我认为查找有关高性能 BufferedImage 主题的信息的最佳位置是 Java 游戏程序员和 Java 游戏 API 程序员社区/论坛。

顺便说一句，请确保您的 BufferedImage 都使用“兼容”模式： TYPE_INT_ARGB 在 Windows 上可能没问题，但在 OS X 等上则不然。因此您想创建它们，执行以下操作：

GraphicsEnvironment.getLocalGraphicsEnvironment().getDefaultScreenDevice().getDefaultConfiguration().createCompatibleImage(width, height, Transparency.TRANSLUCENT);

哎呀德米特定律很痛，谢谢 Java ;)

One relatively fast way, but I don't know if it's the fastest (and I'd like to see other answers) is to have another picture that you never modify and that is always "fully cleared" / "fully transparent" and then you do a raster copy, say you named that copy CLEAR:

imageYouWantToClear.setData( CLEAR.getRaster() );

Note that working with graphics can be very tricky when it comes to performances because there are a lot of not-very-well-documented behavior. For example your images (say the CLEAR one) may be hardware-accelerated but you'd then lose hardware-acceleration as soon as you'd use a mutating method (like say a setRgb()) and it would prove very difficult to realize that you just lost the benefit of hardware acceleration.

I think that the best place to find infos on the subject of performant BufferedImage would be in the Java game-programmers and Java game-API-programmers community/forums.

Btw make sure that both your BufferedImage are using the 'compatible' mode: TYPE_INT_ARGB may be fine on Windows but not on OS X, etc. so you want to create them doing something like:

GraphicsEnvironment.getLocalGraphicsEnvironment().getDefaultScreenDevice().getDefaultConfiguration().createCompatibleImage(width, height, Transparency.TRANSLUCENT);

Ouch the Law-of-Demeter hurts, thanks Java ;)

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