Convert InputStream to byte array in Java

JavaBytearrayInputstream

Java Problem Overview


How do I read an entire InputStream into a byte array?

Java Solutions


Solution 1 - Java

You can use Apache Commons IO to handle this and similar tasks.

The IOUtils type has a static method to read an InputStream and return a byte[].

InputStream is;
byte[] bytes = IOUtils.toByteArray(is);

Internally this creates a ByteArrayOutputStream and copies the bytes to the output, then calls toByteArray(). It handles large files by copying the bytes in blocks of 4KiB.

Solution 2 - Java

You need to read each byte from your InputStream and write it to a ByteArrayOutputStream.

You can then retrieve the underlying byte array by calling toByteArray():

InputStream is = ...
ByteArrayOutputStream buffer = new ByteArrayOutputStream();

int nRead;
byte[] data = new byte[16384];

while ((nRead = is.read(data, 0, data.length)) != -1) {
  buffer.write(data, 0, nRead);
}

return buffer.toByteArray();

Solution 3 - Java

Finally, after twenty years, there’s a simple solution without the need for a 3rd party library, thanks to Java 9:

InputStream is;
…
byte[] array = is.readAllBytes();

Note also the convenience methods readNBytes(byte[] b, int off, int len) and transferTo(OutputStream) addressing recurring needs.

Solution 4 - Java

Use vanilla Java's DataInputStream and its readFully Method (exists since at least Java 1.4):

...
byte[] bytes = new byte[(int) file.length()];
DataInputStream dis = new DataInputStream(new FileInputStream(file));
dis.readFully(bytes);
...

There are some other flavors of this method, but I use this all the time for this use case.

Solution 5 - Java

If you happen to use Google Guava, it'll be as simple as using ByteStreams:

byte[] bytes = ByteStreams.toByteArray(inputStream);

Solution 6 - Java

As always, also Spring framework (spring-core since 3.2.2) has something for you: StreamUtils.copyToByteArray()

Solution 7 - Java

Safe solution (close streams correctly):

  • Java 9+:

     final byte[] bytes;
     try (inputStream) {
     	bytes = inputStream.readAllBytes();
     }
    

  • Java 8:

     public static byte[] readAllBytes(InputStream inputStream) throws IOException {
     	final int bufLen = 4 * 0x400; // 4KB
     	byte[] buf = new byte[bufLen];
     	int readLen;
     	IOException exception = null;
     	
     	try {
     		try (ByteArrayOutputStream outputStream = new ByteArrayOutputStream()) {
     			while ((readLen = inputStream.read(buf, 0, bufLen)) != -1)
     				outputStream.write(buf, 0, readLen);
     			
     			return outputStream.toByteArray();
     		}
     	} catch (IOException e) {
     		exception = e;
     		throw e;
     	} finally {
     		if (exception == null) inputStream.close();
     		else try {
     			inputStream.close();
     		} catch (IOException e) {
     			exception.addSuppressed(e);
     		}
     	}
     }
    

  • Kotlin (when Java 9+ isn't accessible):

     @Throws(IOException::class)
     fun InputStream.readAllBytes(): ByteArray {
         val bufLen = 4 * 0x400 // 4KB
         val buf = ByteArray(bufLen)
         var readLen: Int = 0
    
         ByteArrayOutputStream().use { o ->
             this.use { i ->
                 while (i.read(buf, 0, bufLen).also { readLen = it } != -1)
                     o.write(buf, 0, readLen)
             }
    
             return o.toByteArray()
         }
     }
    

    To avoid nested use see here.


  • Scala (when Java 9+ isn't accessible) (By @Joan. Thx):

    def readAllBytes(inputStream: InputStream): Array[Byte] =
      Stream.continually(inputStream.read).takeWhile(_ != -1).map(_.toByte).toArray
    

Solution 8 - Java

public static byte[] getBytesFromInputStream(InputStream is) throws IOException {
    ByteArrayOutputStream os = new ByteArrayOutputStream();	
    byte[] buffer = new byte[0xFFFF];
    for (int len = is.read(buffer); len != -1; len = is.read(buffer)) {	
	    os.write(buffer, 0, len);
    }
    return os.toByteArray();
}

Solution 9 - Java

In-case someone is still looking for a solution without dependency and If you have a file.

> DataInputStream

 byte[] data = new byte[(int) file.length()];
 DataInputStream dis = new DataInputStream(new FileInputStream(file));
 dis.readFully(data);
 dis.close();

> ByteArrayOutputStream

 InputStream is = new FileInputStream(file);
 ByteArrayOutputStream buffer = new ByteArrayOutputStream();
 int nRead;
 byte[] data = new byte[(int) file.length()];
 while ((nRead = is.read(data, 0, data.length)) != -1) {
     buffer.write(data, 0, nRead);
 }

> RandomAccessFile

 RandomAccessFile raf = new RandomAccessFile(file, "r");
 byte[] data = new byte[(int) raf.length()];
 raf.readFully(data);

Solution 10 - Java

Do you really need the image as a byte[]? What exactly do you expect in the byte[] - the complete content of an image file, encoded in whatever format the image file is in, or RGB pixel values?

Other answers here show you how to read a file into a byte[]. Your byte[] will contain the exact contents of the file, and you'd need to decode that to do anything with the image data.

Java's standard API for reading (and writing) images is the ImageIO API, which you can find in the package javax.imageio. You can read in an image from a file with just a single line of code:

BufferedImage image = ImageIO.read(new File("image.jpg"));

This will give you a BufferedImage, not a byte[]. To get at the image data, you can call getRaster() on the BufferedImage. This will give you a Raster object, which has methods to access the pixel data (it has several getPixel() / getPixels() methods).

Lookup the API documentation for javax.imageio.ImageIO, java.awt.image.BufferedImage, java.awt.image.Raster etc.

ImageIO supports a number of image formats by default: JPEG, PNG, BMP, WBMP and GIF. It's possible to add support for more formats (you'd need a plug-in that implements the ImageIO service provider interface).

See also the following tutorial: Working with Images

Solution 11 - Java

If you don't want to use the Apache commons-io library, this snippet is taken from the sun.misc.IOUtils class. It's nearly twice as fast as the common implementation using ByteBuffers:

public static byte[] readFully(InputStream is, int length, boolean readAll)
        throws IOException {
    byte[] output = {};
    if (length == -1) length = Integer.MAX_VALUE;
    int pos = 0;
    while (pos < length) {
        int bytesToRead;
        if (pos >= output.length) { // Only expand when there's no room
            bytesToRead = Math.min(length - pos, output.length + 1024);
            if (output.length < pos + bytesToRead) {
                output = Arrays.copyOf(output, pos + bytesToRead);
            }
        } else {
            bytesToRead = output.length - pos;
        }
        int cc = is.read(output, pos, bytesToRead);
        if (cc < 0) {
            if (readAll && length != Integer.MAX_VALUE) {
                throw new EOFException("Detect premature EOF");
            } else {
                if (output.length != pos) {
                    output = Arrays.copyOf(output, pos);
                }
                break;
            }
        }
        pos += cc;
    }
    return output;
}

Solution 12 - Java

ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buffer = new byte[1024];
while (true) {
	int r = in.read(buffer);
	if (r == -1) break;
	out.write(buffer, 0, r);
}

byte[] ret = out.toByteArray();

Solution 13 - Java

Input Stream is ...
ByteArrayOutputStream bos = new ByteArrayOutputStream();
int next = in.read();
while (next > -1) {
    bos.write(next);
    next = in.read();
}
bos.flush();
byte[] result = bos.toByteArray();
bos.close();

Solution 14 - Java

@Adamski: You can avoid buffer entirely.

Code copied from http://www.exampledepot.com/egs/java.io/File2ByteArray.html (Yes, it is very verbose, but needs half the size of memory as the other solution.)

// Returns the contents of the file in a byte array.
public static byte[] getBytesFromFile(File file) throws IOException {
    InputStream is = new FileInputStream(file);

    // Get the size of the file
    long length = file.length();

    // You cannot create an array using a long type.
    // It needs to be an int type.
    // Before converting to an int type, check
    // to ensure that file is not larger than Integer.MAX_VALUE.
    if (length > Integer.MAX_VALUE) {
        // File is too large
    }

    // Create the byte array to hold the data
    byte[] bytes = new byte[(int)length];

    // Read in the bytes
    int offset = 0;
    int numRead = 0;
    while (offset < bytes.length
           && (numRead=is.read(bytes, offset, bytes.length-offset)) >= 0) {
        offset += numRead;
    }

    // Ensure all the bytes have been read in
    if (offset < bytes.length) {
        throw new IOException("Could not completely read file "+file.getName());
    }

    // Close the input stream and return bytes
    is.close();
    return bytes;
}

Solution 15 - Java

Java 9 will give you finally a nice method:

InputStream in = ...;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
in.transferTo( bos );
byte[] bytes = bos.toByteArray();

Solution 16 - Java

We are seeing some delay for few AWS transaction, while converting S3 object to ByteArray.

Note: S3 Object is PDF document (max size is 3 mb).

We are using the option #1 (org.apache.commons.io.IOUtils) to convert the S3 object to ByteArray. We have noticed S3 provide the inbuild IOUtils method to convert the S3 object to ByteArray, we are request you to confirm what is the best way to convert the S3 object to ByteArray to avoid the delay.

Option #1:

import org.apache.commons.io.IOUtils;
is = s3object.getObjectContent();
content =IOUtils.toByteArray(is);

Option #2:

import com.amazonaws.util.IOUtils;
is = s3object.getObjectContent();
content =IOUtils.toByteArray(is);

Also let me know if we have any other better way to convert the s3 object to bytearray

Solution 17 - Java

I know it's too late but here I think is cleaner solution that's more readable...

/**
 * method converts {@link InputStream} Object into byte[] array.
 * 
 * @param stream the {@link InputStream} Object.
 * @return the byte[] array representation of received {@link InputStream} Object.
 * @throws IOException if an error occurs.
 */
public static byte[] streamToByteArray(InputStream stream) throws IOException {

    byte[] buffer = new byte[1024];
    ByteArrayOutputStream os = new ByteArrayOutputStream();

    int line = 0;
    // read bytes from stream, and store them in buffer
    while ((line = stream.read(buffer)) != -1) {
        // Writes bytes from byte array (buffer) into output stream.
        os.write(buffer, 0, line);
    }
    stream.close();
    os.flush();
    os.close();
    return os.toByteArray();
}

Solution 18 - Java

I tried to edit @numan's answer with a fix for writing garbage data but edit was rejected. While this short piece of code is nothing brilliant I can't see any other better answer. Here's what makes most sense to me:

ByteArrayOutputStream out = new ByteArrayOutputStream();
byte[] buffer = new byte[1024]; // you can configure the buffer size
int length;

while ((length = in.read(buffer)) != -1) out.write(buffer, 0, length); //copy streams
in.close(); // call this in a finally block

byte[] result = out.toByteArray();

btw ByteArrayOutputStream need not be closed. try/finally constructs omitted for readability

Solution 19 - Java

See the InputStream.available() documentation:

> It is particularly important to realize that you must not use this > method to size a container and assume that you can read the entirety > of the stream without needing to resize the container. Such callers > should probably write everything they read to a ByteArrayOutputStream > and convert that to a byte array. Alternatively, if you're reading > from a file, File.length returns the current length of the file > (though assuming the file's length can't change may be incorrect, > reading a file is inherently racy).

Solution 20 - Java

Wrap it in a DataInputStream if that is off the table for some reason, just use read to hammer on it until it gives you a -1 or the entire block you asked for.

public int readFully(InputStream in, byte[] data) throws IOException {
    int offset = 0;
    int bytesRead;
    boolean read = false;
    while ((bytesRead = in.read(data, offset, data.length - offset)) != -1) {
        read = true;
        offset += bytesRead;
        if (offset >= data.length) {
            break;
        }
    }
    return (read) ? offset : -1;
}

Solution 21 - Java

Java 8 way (thanks to BufferedReader and Adam Bien)

private static byte[] readFully(InputStream input) throws IOException {
    try (BufferedReader buffer = new BufferedReader(new InputStreamReader(input))) {
        return buffer.lines().collect(Collectors.joining("\n")).getBytes(<charset_can_be_specified>);
    }
}

Note that this solution wipes carriage return ('\r') and can be inappropriate.

Solution 22 - Java

Here is an optimized version, that tries to avoid copying data bytes as much as possible:

private static byte[] loadStream (InputStream stream) throws IOException {
   int available = stream.available();
   int expectedSize = available > 0 ? available : -1;
   return loadStream(stream, expectedSize);
}

private static byte[] loadStream (InputStream stream, int expectedSize) throws IOException {
   int basicBufferSize = 0x4000;
   int initialBufferSize = (expectedSize >= 0) ? expectedSize : basicBufferSize;
   byte[] buf = new byte[initialBufferSize];
   int pos = 0;
   while (true) {
      if (pos == buf.length) {
         int readAhead = -1;
         if (pos == expectedSize) {
            readAhead = stream.read();       // test whether EOF is at expectedSize
            if (readAhead == -1) {
               return buf;
            }
         }
         int newBufferSize = Math.max(2 * buf.length, basicBufferSize);
         buf = Arrays.copyOf(buf, newBufferSize);
         if (readAhead != -1) {
            buf[pos++] = (byte)readAhead;
         }
      }
      int len = stream.read(buf, pos, buf.length - pos);
      if (len < 0) {
         return Arrays.copyOf(buf, pos);
      }
      pos += len;
   }
}

Solution 23 - Java

The other case to get correct byte array via stream, after send request to server and waiting for the response.

/**
         * Begin setup TCP connection to PC app
         * to open integrate connection between mobile app and pc app (or mobile app)
         */
        mSocket = new Socket(IP, port);
       // mSocket.setSoTimeout(30000);

        DataOutputStream mDos = new DataOutputStream(mSocket.getOutputStream());

        String str = "MobileRequest#" + params[0] + "#<EOF>";

        mDos.write(str.getBytes());

        try {
            Thread.sleep(1000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }

        /* Since data are accepted as byte, all of them will be collected in the
        following byte array which initialised with accepted data length. */
        DataInputStream mDis = new DataInputStream(mSocket.getInputStream());
        byte[] data = new byte[mDis.available()];

        // Collecting data into byte array
        for (int i = 0; i < data.length; i++)
            data[i] = mDis.readByte();

        // Converting collected data in byte array into String.
        String RESPONSE = new String(data);

Solution 24 - Java

You're doing an extra copy if you use ByteArrayOutputStream. If you know the length of the stream before you start reading it (e.g. the InputStream is actually a FileInputStream, and you can call file.length() on the file, or the InputStream is a zipfile entry InputStream, and you can call zipEntry.length()), then it's far better to write directly into the byte[] array -- it uses half the memory, and saves time.

// Read the file contents into a byte[] array
byte[] buf = new byte[inputStreamLength];
int bytesRead = Math.max(0, inputStream.read(buf));

// If needed: for safety, truncate the array if the file may somehow get
// truncated during the read operation
byte[] contents = bytesRead == inputStreamLength ? buf
                  : Arrays.copyOf(buf, bytesRead);

N.B. the last line above deals with files getting truncated while the stream is being read, if you need to handle that possibility, but if the file gets longer while the stream is being read, the contents in the byte[] array will not be lengthened to include the new file content, the array will simply be truncated to the old length inputStreamLength.

Solution 25 - Java

I use this.

public static byte[] toByteArray(InputStream is) throws IOException {
        ByteArrayOutputStream output = new ByteArrayOutputStream();
        try {
            byte[] b = new byte[4096];
            int n = 0;
            while ((n = is.read(b)) != -1) {
                output.write(b, 0, n);
            }
            return output.toByteArray();
        } finally {
            output.close();
        }
    }

Solution 26 - Java

This is my copy-paste version:

@SuppressWarnings("empty-statement")
public static byte[] inputStreamToByte(InputStream is) throws IOException {
    if (is == null) {
        return null;
    }
    // Define a size if you have an idea of it.
    ByteArrayOutputStream r = new ByteArrayOutputStream(2048);
    byte[] read = new byte[512]; // Your buffer size.
    for (int i; -1 != (i = is.read(read)); r.write(read, 0, i));
    is.close();
    return r.toByteArray();
}

Solution 27 - Java

Java 7 and later:

import sun.misc.IOUtils;
...
InputStream in = ...;
byte[] buf = IOUtils.readFully(in, -1, false);

Solution 28 - Java

You can try Cactoos:

byte[] array = new BytesOf(stream).bytes();

Solution 29 - Java

Solution in Kotlin (will work in Java too, of course), which includes both cases of when you know the size or not:

    fun InputStream.readBytesWithSize(size: Long): ByteArray? {
        return when {
            size < 0L -> this.readBytes()
            size == 0L -> ByteArray(0)
            size > Int.MAX_VALUE -> null
            else -> {
                val sizeInt = size.toInt()
                val result = ByteArray(sizeInt)
                readBytesIntoByteArray(result, sizeInt)
                result
            }
        }
    }

    fun InputStream.readBytesIntoByteArray(byteArray: ByteArray,bytesToRead:Int=byteArray.size) {
        var offset = 0
        while (true) {
            val read = this.read(byteArray, offset, bytesToRead - offset)
            if (read == -1)
                break
            offset += read
            if (offset >= bytesToRead)
                break
        }
    }

If you know the size, it saves you on having double the memory used compared to the other solutions (in a brief moment, but still could be useful). That's because you have to read the entire stream to the end, and then convert it to a byte array (similar to ArrayList which you convert to just an array).

So, if you are on Android, for example, and you got some Uri to handle, you can try to get the size using this:

    fun getStreamLengthFromUri(context: Context, uri: Uri): Long {
        context.contentResolver.query(uri, arrayOf(MediaStore.MediaColumns.SIZE), null, null, null)?.use {
            if (!it.moveToNext())
                return@use
            val fileSize = it.getLong(it.getColumnIndex(MediaStore.MediaColumns.SIZE))
            if (fileSize > 0)
                return fileSize
        }
        //if you wish, you can also get the file-path from the uri here, and then try to get its size, using this: https://stackoverflow.com/a/61835665/878126
        FileUtilEx.getFilePathFromUri(context, uri, false)?.use {
            val file = it.file
            val fileSize = file.length()
            if (fileSize > 0)
                return fileSize
        }
        context.contentResolver.openInputStream(uri)?.use { inputStream ->
            if (inputStream is FileInputStream)
                return inputStream.channel.size()
            else {
                var bytesCount = 0L
                while (true) {
                    val available = inputStream.available()
                    if (available == 0)
                        break
                    val skip = inputStream.skip(available.toLong())
                    if (skip < 0)
                        break
                    bytesCount += skip
                }
                if (bytesCount > 0L)
                    return bytesCount
            }
        }
        return -1L
    }

Solution 30 - Java

You can use cactoos library with provides reusable object-oriented Java components. OOP is emphasized by this library, so no static methods, NULLs, and so on, only real objects and their contracts (interfaces). A simple operation like reading InputStream, can be performed like that

final InputStream input = ...;
final Bytes bytes = new BytesOf(input);
final byte[] array = bytes.asBytes();
Assert.assertArrayEquals(
    array,
    new byte[]{65, 66, 67}
);

Having a dedicated type Bytes for working with data structure byte[] enables us to use OOP tactics for solving tasks at hand. Something that a procedural "utility" method will forbid us to do. For example, you need to enconde bytes you've read from this InputStream to Base64. In this case you will use Decorator pattern and wrap Bytes object within implementation for Base64. cactoos already provides such implementation:

final Bytes encoded = new BytesBase64(
    new BytesOf(
        new InputStreamOf("XYZ")
    )
);
Assert.assertEquals(new TextOf(encoded).asString(), "WFla");

You can decode them in the same manner, by using Decorator pattern

final Bytes decoded = new Base64Bytes(
    new BytesBase64(
        new BytesOf(
            new InputStreamOf("XYZ")
        )
    )
);
Assert.assertEquals(new TextOf(decoded).asString(), "XYZ");

Whatever your task is you will be able to create own implementation of Bytes to solve it.

Solution 31 - Java

/*InputStream class_InputStream = null;
I am reading class from DB 
class_InputStream = rs.getBinaryStream(1);
Your Input stream could be from any source
*/
int thisLine;
ByteArrayOutputStream bos = new ByteArrayOutputStream();
while ((thisLine = class_InputStream.read()) != -1) {
	bos.write(thisLine);
}
bos.flush();
byte [] yourBytes = bos.toByteArray();

/*Don't forget in the finally block to close ByteArrayOutputStream & InputStream
 In my case the IS is from resultset so just closing the rs will do it*/

if (bos != null){
    bos.close();
}

Solution 32 - Java

This works for me,

if(inputStream != null){
                ByteArrayOutputStream contentStream = readSourceContent(inputStream);
                String stringContent = contentStream.toString();
                byte[] byteArr = encodeString(stringContent);
            }

readSourceContent()

public static ByteArrayOutputStream readSourceContent(InputStream inputStream) throws IOException {
		ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
		int nextChar;
		try {
			while ((nextChar = inputStream.read()) != -1) {
				outputStream.write(nextChar);
			}
			outputStream.flush();
		} catch (IOException e) {
			throw new IOException("Exception occurred while reading content", e);
		}

		return outputStream;
	}

encodeString()

public static byte[] encodeString(String content) throws UnsupportedEncodingException {
        byte[] bytes;
        try {
            bytes = content.getBytes();

        } catch (UnsupportedEncodingException e) {
            String msg = ENCODING + " is unsupported encoding type";
            log.error(msg,e);
            throw new UnsupportedEncodingException(msg, e);
        }
        return bytes;
    }

Solution 33 - Java

Below Codes

public static byte[] serializeObj(Object obj) throws IOException {
  ByteArrayOutputStream baOStream = new ByteArrayOutputStream();
  ObjectOutputStream objOStream = new ObjectOutputStream(baOStream);

  objOStream.writeObject(obj); 
  objOStream.flush();
  objOStream.close();
  return baOStream.toByteArray(); 
} 

OR

BufferedImage img = ...
ByteArrayOutputStream baos = new ByteArrayOutputStream(1000);
ImageIO.write(img, "jpeg", baos);
baos.flush();
byte[] result = baos.toByteArray();
baos.close();

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