Java- Convert bufferedimage to byte[] without writing to disk

Question

I'm trying to send multiple images over a socket using java but I need a faster way to convert the images to a byte array so I can send them. I tried the following code but it wrote about 10,000 images to my C:\ drive. Is there a way to make this conversion without writing to disk? Thanks!

ByteArrayOutputStream outputStream = new ByteArrayOutputStream();

                    //ImageIO.setUseCache(false);
                    ImageIO.write(bi.getImage(), "jpg", outputStream);

                    byte[] imageBytes = outputStream.toByteArray();

Note that a little extra time doing the image compression using a higher compression (lower quality) JPEG can save a lot of network time. I think you need to consider the byte size actually sent, which will be significantly larger with a pure BufferedImage. BTW - what is the source of these images? If it is screenshots for e.g., often PNG can provide a smaller byte size than a default compression JPEG. If it were continuous screenshots, perhaps even a video stream. — Andrew Thompson
– Andrew Thompson, Commented Apr 20, 2012 at 13:47
The source is continuous screenshots that I'm essentially making into a stream. But yes, I need to figure out how to compress the images without writing them to disk. That is my ultimate goal — tier1
– tier1, Commented Apr 20, 2012 at 15:08

stacker · Accepted Answer · 2012-04-20 13:26:24Z

52

This should work:

byte[] imageBytes = ((DataBufferByte) bufferedImage.getData().getDataBuffer()).getData();

answered Apr 20, 2012 at 13:26

stacker

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4 Comments

Sam Barnum Over a year ago

This will create a copy of the image. If you just want a direct reference to the bytes, call bufferedImage.getRaster().getDataBuffer() and cast it accordingly (it's not always safe to cast to a DataBufferByte)

user3712476 Over a year ago

this doesn't seem to work for me. I get java.awt.image.DataBufferInt cannot be cast to java.awt.image.DataBufferByte my bufferedimage is coming from a Robot.createScreenCapture() not sure if that matters

stacker Over a year ago

@user3712476 seems that Robot isn't a class from JRE libs, you could try to assign it to variable of type DataBufferInt and cast it to DataBufferInt .

Surabhi Mundra Over a year ago

bufferedImage.getData().getDataBuffer() is creating invalid byte array for buffered image

EderBaum · Accepted Answer · 2016-04-07 14:46:51Z

4

The code below it's really fast (few milliseconds)

import com.sun.image.codec.jpeg.JPEGCodec;
import com.sun.image.codec.jpeg.JPEGImageEncoder;

public byte[] toByteArray(BufferedImage image) throws IOException {
    ByteArrayOutputStream baos = new ByteArrayOutputStream();            
    JPEGImageEncoder encoder = JPEGCodec.createJPEGEncoder(baos);
    encoder.encode(image);            
    return baos.toByteArray();
}

answered Apr 7, 2016 at 14:46

EderBaum

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3 Comments

Erieze Lagera Over a year ago

It's better to stay away with com.sun package.

Serġan Over a year ago

Official post on why sun package should not be used: oracle.com/technetwork/java/faq-sun-packages-142232.html

meawoppl Over a year ago

Not an answer to the question.

bugCracker · Accepted Answer · 2018-07-20 04:13:06Z

4

ByteArrayOutputStream baos;
ImageIO.write(bufferedImage, "png", baos);
byte[] imageBytes = baos.toByteArray();

answered Jul 20, 2018 at 4:13

bugCracker

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Comments

soulcheck · Accepted Answer · 2012-04-20 13:28:29Z

2

Try using:

ImageIO.setUseCache(false);

Before writing, maybe that helps.

answered Apr 20, 2012 at 13:28

soulcheck

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4 Comments

tier1 Over a year ago

I tried this, however, I believe this still writes to disk, it just deletes it after the operation is complete.

soulcheck Over a year ago

it shouldn't, at least according to docs :

Setting this flag to false disallows the use of disk for future streams, which may be advantageous when working with small images, as the overhead of creating and destroying files is removed.

. Maybe your program writes to disk somewhere else?

tier1 Over a year ago

Yeah, I saw that in the docs as well. Either way, the operation is painfully slow for what I'm looking to do. Thanks though.

Alexander Klimetschek Over a year ago

AFAIU the setUseCache() flag only applies to input streams when reading an image from a stream. ImageIO either loads it into memory or moves it to a disk (when use cache = true), to support all kinds of random access operations normally not possible with InputStream. But the file format parsers (who operate on that ImageInputStream) will always create a BufferedImage as result, which will be completely in memory and uncompressed. Only Java Advanced Imaging has parsers to support tile based reading and conversion, without ever having to store the entire image in memory.

Zaz Gmy · Accepted Answer · 2012-04-20 13:28:44Z

1

BufferedImage img = ImageIO.read(new ByteArrayInputStream(bytes));
byte[] bytes = new byte[buf.capacity()];
buf.get(bytes, 0, bytes.length);

answered Apr 20, 2012 at 13:28

Zaz Gmy

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Phani · Accepted Answer · 2012-04-20 13:26:52Z

0

Use Apache Commons IO Utils Apache Commons

IOUtils.copy(inputStream,outputStream);

IO Utils API supports the large buffers easily

answered Apr 20, 2012 at 13:26

Phani

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