使用这个bitmap.recycle();这有助于没有任何图像质量问题。

在我的一个应用程序中，我需要从相机/画廊中拍照。如果用户从相机点击图像(可能是200万，500万或800万)，图像大小从kBs到mb不等。如果图像大小较小(或高达1-2MB)以上代码工作正常，但如果我有图像的大小超过4MB或5MB，然后OOM进入帧:(

然后我努力解决这个问题，最后我对Fedor的(所有功劳都归功于Fedor，因为它是一个很好的解决方案)代码做出了以下改进:)

private Bitmap decodeFile(String fPath) {
    // Decode image size
    BitmapFactory.Options opts = new BitmapFactory.Options();
    /*
     * If set to true, the decoder will return null (no bitmap), but the
     * out... fields will still be set, allowing the caller to query the
     * bitmap without having to allocate the memory for its pixels.
     */
    opts.inJustDecodeBounds = true;
    opts.inDither = false; // Disable Dithering mode
    opts.inPurgeable = true; // Tell to gc that whether it needs free
                                // memory, the Bitmap can be cleared
    opts.inInputShareable = true; // Which kind of reference will be used to
                                    // recover the Bitmap data after being
                                    // clear, when it will be used in the
                                    // future

    BitmapFactory.decodeFile(fPath, opts);

    // The new size we want to scale to
    final int REQUIRED_SIZE = 70;

    // Find the correct scale value. 
    int scale = 1;

    if (opts.outHeight > REQUIRED_SIZE || opts.outWidth > REQUIRED_SIZE) {

        // Calculate ratios of height and width to requested height and width
        final int heightRatio = Math.round((float) opts.outHeight
                / (float) REQUIRED_SIZE);
        final int widthRatio = Math.round((float) opts.outWidth
                / (float) REQUIRED_SIZE);

        // Choose the smallest ratio as inSampleSize value, this will guarantee
        // a final image with both dimensions larger than or equal to the
        // requested height and width.
        scale = heightRatio < widthRatio ? heightRatio : widthRatio;//
    }

    // Decode bitmap with inSampleSize set
    opts.inJustDecodeBounds = false;

    opts.inSampleSize = scale;

    Bitmap bm = BitmapFactory.decodeFile(fPath, opts).copy(
            Bitmap.Config.RGB_565, false);

    return bm;

}

希望这篇文章能对面临同样问题的朋友有所帮助!

欲了解更多，请参考此

这对我很管用!

public Bitmap readAssetsBitmap(String filename) throws IOException {
    try {
        BitmapFactory.Options options = new BitmapFactory.Options(); 
        options.inPurgeable = true;
        Bitmap bitmap = BitmapFactory.decodeStream(assets.open(filename), null, options);
        if(bitmap == null) {
            throw new IOException("File cannot be opened: It's value is null");
        } else {
            return bitmap;
        }
    } catch (IOException e) {
        throw new IOException("File cannot be opened: " + e.getMessage());
    }
}

这将获得适当的位图并减少内存消耗

JAVA

Bitmap bm = null;

BitmapFactory.Options bmpOption = new BitmapFactory.Options();
bmpOption.inJustDecodeBounds = true;

FileInputStream fis = new FileInputStream(file);
BitmapFactory.decodeStream(fis, null, bmpOption);
fis.close();

int scale = 1;

if (bmpOption.outHeight > IMAGE_MAX_SIZE || bmpOption.outWidth > IMAGE_MAX_SIZE) {
    scale = (int)Math.pow(2, (int) Math.ceil(Math.log(IMAGE_MAX_SIZE / 
       (double) Math.max(bmpOption.outHeight, bmpOption.outWidth)) / Math.log(0.5)));
}

BitmapFactory.Options bmpOption2 = new BitmapFactory.Options();
bmpOption2.inSampleSize = scale;
fis = new FileInputStream(file);
bm = BitmapFactory.decodeStream(fis, null, bmpOption2);
fis.close();

科特林

val bm:Bitmap = null
val bmpOption = BitmapFactory.Options()
bmpOption.inJustDecodeBounds = true
val fis = FileInputStream(file)
BitmapFactory.decodeStream(fis, null, bmpOption)
fis.close()
val scale = 1
if (bmpOption.outHeight > IMAGE_MAX_SIZE || bmpOption.outWidth > IMAGE_MAX_SIZE)
{
  scale = Math.pow(2.0, Math.ceil((Math.log((IMAGE_MAX_SIZE / Math.max(bmpOption.outHeight, bmpOption.outWidth) as Double)) / Math.log(0.5))).toInt().toDouble()).toInt()
}
val bmpOption2 = BitmapFactory.Options()
bmpOption2.inSampleSize = scale
fis = FileInputStream(file)
bm = BitmapFactory.decodeStream(fis, null, bmpOption2)
fis.close()

我有同样的问题，并通过避免BitmapFactory.decodeStream或decodeFile函数来解决它，而是使用BitmapFactory.decodeFileDescriptor

decodeFileDescriptor看起来像调用了不同于decodeStream/decodeFile的本地方法。

不管怎样，这样做是有效的(注意，我添加了一些选项，就像上面的一些选项一样，但这并不是造成不同的原因。关键是调用BitmapFactory.decodeFileDescriptor而不是decodeStream或decodeFile):

private void showImage(String path)   {

    Log.i("showImage","loading:"+path);
    BitmapFactory.Options bfOptions=new BitmapFactory.Options();
    bfOptions.inDither=false;                     //Disable Dithering mode
    bfOptions.inPurgeable=true;                   //Tell to gc that whether it needs free memory, the Bitmap can be cleared
    bfOptions.inInputShareable=true;              //Which kind of reference will be used to recover the Bitmap data after being clear, when it will be used in the future
    bfOptions.inTempStorage=new byte[32 * 1024]; 

    File file=new File(path);
    FileInputStream fs=null;
    try {
        fs = new FileInputStream(file);
    } catch (FileNotFoundException e) {
        //TODO do something intelligent
        e.printStackTrace();
    }

    try {
        if(fs!=null) bm=BitmapFactory.decodeFileDescriptor(fs.getFD(), null, bfOptions);
    } catch (IOException e) {
        //TODO do something intelligent
        e.printStackTrace();
    } finally{ 
        if(fs!=null) {
            try {
                fs.close();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        }
    }
    //bm=BitmapFactory.decodeFile(path, bfOptions); This one causes error: java.lang.OutOfMemoryError: bitmap size exceeds VM budget

    im.setImageBitmap(bm);
    //bm.recycle();
    bm=null;                        
}

我认为在decodeStream/decodeFile中使用的本机函数有问题。我已经确认使用decodeFileDescriptor时调用了不同的本机方法。此外，我读到的是“图像(位图)不是以标准的Java方式分配的，而是通过本机调用;分配是在虚拟堆之外完成的，但是 不算数!”

这段代码将有助于从可绘制的图中加载大的位图

public class BitmapUtilsTask extends AsyncTask<Object, Void, Bitmap> {

    Context context;

    public BitmapUtilsTask(Context context) {
        this.context = context;
    }

    /**
     * Loads a bitmap from the specified url.
     * 
     * @param url The location of the bitmap asset
     * @return The bitmap, or null if it could not be loaded
     * @throws IOException
     * @throws MalformedURLException
     */
    public Bitmap getBitmap() throws MalformedURLException, IOException {       

        // Get the source image's dimensions
        int desiredWidth = 1000;
        BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;

        BitmapFactory.decodeResource(context.getResources(), R.drawable.green_background , options);

        int srcWidth = options.outWidth;
        int srcHeight = options.outHeight;

        // Only scale if the source is big enough. This code is just trying
        // to fit a image into a certain width.
        if (desiredWidth > srcWidth)
            desiredWidth = srcWidth;

        // Calculate the correct inSampleSize/scale value. This helps reduce
        // memory use. It should be a power of 2
        int inSampleSize = 1;
        while (srcWidth / 2 > desiredWidth) {
            srcWidth /= 2;
            srcHeight /= 2;
            inSampleSize *= 2;
        }
        // Decode with inSampleSize
        options.inJustDecodeBounds = false;
        options.inDither = false;
        options.inSampleSize = inSampleSize;
        options.inScaled = false;
        options.inPreferredConfig = Bitmap.Config.ARGB_8888;
        options.inPurgeable = true;
        Bitmap sampledSrcBitmap;

        sampledSrcBitmap =  BitmapFactory.decodeResource(context.getResources(), R.drawable.green_background , options);

        return sampledSrcBitmap;
    }

    /**
     * The system calls this to perform work in a worker thread and delivers
     * it the parameters given to AsyncTask.execute()
     */
    @Override
    protected Bitmap doInBackground(Object... item) {
        try { 
          return getBitmap();
        } catch (MalformedURLException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }
}