如何从输入流中旋转图像而不将其加载到内存中？

Question

我浏览了这些主题，但仍然没有找到以下问题的答案。

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我有一个Uri，它代表一个大图像。我需要将此图像的旋转副本保存到 SD 卡。但是当我这样做时

BitmapFactory.decodeStream(contentResolver.openInputStream(uri));

我收到 OutOfMemoryError。

我想做的是这样的

    ContentResolver resolver = getContentResolver();
    InputStream inputStream = null;
    FileOutputStream outputStream = null;
    File imageFile = null;
    try {
        inputStream = resolver.openInputStream(uri);


        //I don't know how to apply this matrix to the bytes of InputStream
        Matrix matrix = new Matrix();
        matrix.postRotate(90);

        imageFile = createImageFile();

        outputStream = new FileOutputStream(imageFile);
        byte[] buffer = new byte[1024];
        while (inputStream.read(buffer) != -1) {

            //I'd like to apply the 'matrix' to 'buffer' here somehow but I can't
            //since 'buffer' contains the bytes of the image, not the pixel values of the bitmap
            outputStream.write(buffer);
        }
    } catch (IOException e) {
        throw new RuntimeException(e);
    } finally {
        Util.closeQuietly(inputStream);
        Util.closeQuietly(outputStream);
    }

如果有的话，有什么解决办法？

Answer 1

如果您可以接受较低质量的图像，您可以使用 BitmapFactory.Options 读取它 使用这些选项时，您通常无法直观地分辨出区别

 BitmapFactory.Options options = new BitmapFactory.Options();
 options.inSampleSize = 2; //Scale it down
 options.inPreferredConfig = Bitmap.Config.RBG_565; // Less memory intensive color format
 Bitmap bitmap = BitmapFactory.decodeStream( fis, null, options );

希望这会有所帮助。

如果不先阅读它，我不明白如何操作它。