android - 存储位图的最佳方式

Question

现在我有一些类负责保存内部内存位图，然后在 ListViews 中列出。

列出后，诉诸一种方法，告诉您要返回的图像名称并接收相应的位图。

当应用程序启动时，将您需要的所有图像加载到内存（HashMap）中，一旦将图像从内部存储器直接加载到列表视图，该过程需要很长时间。

我正在使用视图的回收，每个项目都有两个和两个图像视图文本视图。但是，滚动非常慢并且有很多滞后。

将图像保存在内存中（哈希图）是一个很好的解决方案吗？对于在列表时加载到内存中的速度更慢。图像有 100x100 像素。

我的问题是：我的延迟问题可能与图像都在内存中这一事实有关？什么是最好的解决方案？

编辑：

07-25 16:56:33.195: V/ADAPTER(8648): recicling view [1]
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每当我滚动（列表继续）时，我都会得到这个输出。其中 [] 内的内容对应位置。他不断地显示0-6的位置是不是很奇怪？

编辑

现在发现了一个可能的原因。

我目前是一个 tab_activity，其中有所有必要的视图。在 xml 中添加了一个 tab_activity top_bar，其中我有一个不断更改的图像。当图像处于自动模式（不断更改）时，选项卡内的 ListViews 被“重新绘制”，因此存在滞后。

我正在使用一项服务（音乐），它会在 x 时间内改变 x 的图像。

为什么我查看重绘图像时要重绘 ListView？

score 0 · Accepted Answer

I'm not sure if the loading of the images is causing the lag, yet if it is, then I suggest you create some sort of UI that reports to the user that the images are loading. (To check wether the images are causing the lag, I suggest you have a look at DDMS are simply log with the LogCat)

I use the following code to save bitmaps to the memory:

The following will write to the memory:

public void writeBitmapToMemory(String filename, Bitmap bitmap) {
        FileOutputStream fos;
        // Use the compress method on the Bitmap object to write image to the OutputStream
        try {
            fos = game.openFileOutput(filename, Context.MODE_PRIVATE);
            // Writing the bitmap to the output stream
            bitmap.compress(Bitmap.CompressFormat.PNG, 100, fos);
            fos.close();
            // this.gameEngineLog.d(classTAG, "Bitmap successfully written: " + filename);
        } 
        catch (FileNotFoundException e) {
            e.printStackTrace();
            // this.gameEngineLog.d(classTAG, "Bitmap couldn't be written: " + filename);

        } 
        catch (IOException e) {
            e.printStackTrace();
            // this.gameEngineLog.d(classTAG, "Bitmap couldn't be written: " + filename);

        }

    }

this will read from the memory:

public Bitmap readBitmapFromMemory(String filename) {
        Bitmap defautBitmap = null;
        File filePath = game.getFileStreamPath(filename);
        FileInputStream fi;
        try {
            fi = new FileInputStream(filePath);
            defautBitmap = BitmapFactory.decodeStream(fi);
            // this.gameEngineLog.d(classTAG, "Bitmap successfully read: " + filename);
        } 
        catch (FileNotFoundException e) {
            e.printStackTrace();
            // this.gameEngineLog.d(classTAG, "Bitmap couldn't be opened: " + filename);
        }

        return defautBitmap;

    }

android - 存储位图的最佳方式

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