In Java 8 how do I go from an InputStream to a byte array compressing the stream with zip along the way?

Question

In other words I have:

InputStream inputStream = getInputStreamFromSource();
byte[] output = zipOutputStreamAndConvertToByteArray(inputStream);

How would the function zipOutputStreamAndConvertToByteArray be implemented?

private byte[] zipOutputStreamAndConvertToByteArray(InputStream inputStream)
{
    // what code goes here?
}

Answer 1

This creates a zip file with file named file:

private static byte[] zipOutputStreamAndConvertToByteArray(InputStream inputStream) throws IOException {
    ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
    try (ZipOutputStream zip = new ZipOutputStream(outputStream)) {
        zip.putNextEntry(new ZipEntry("file"));
        try (InputStream in = inputStream) { 
        // this try block can be replaced with IOUtils.copy or ByteStreams.copy
            byte[] buffer = new byte[4096];
            int len;
            while ((len = in.read(buffer)) > 0) {
                zip.write(buffer, 0, len);
            }
        }
        zip.closeEntry();
    }

    return outputStream.toByteArray();
}

Answer 2

Note that Java 9 allows to simplify the solution of Adam Siemion’s answer significantly:

private static byte[] zipIntoByteArray(InputStream inputStream) throws IOException {
    ByteArrayOutputStream outputStream = new ByteArrayOutputStream();
    try(ZipOutputStream zip = new ZipOutputStream(outputStream); inputStream) {
        zip.putNextEntry(new ZipEntry("file"));
        inputStream.transferTo(zip);
    }
    return outputStream.toByteArray();
}

The first changes are not Java 9 specific, you can manage both streams with a single try statement and you don’t need to close the last (sole) zip entry as it will get closed automatically when the stream is closed. Then, Java 9 allows you to specify an existing variable like inputStream without declaring another variable in the try statement. Further, you can use transferTo to transfer all remaining data from an InputStream to an OutputStream, without the need to implement a copying routine.

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As said by Izruo in this comment when you don’t need the data in the zip file format and only want to hold a single file, you can get rid of the zip file specific overhead by using the DEFLATE algorithm directly. We could do this similar to the solution above, just replacing ZipOutputStream with DeflaterOutputStream, but there’s another alternative, using DeflaterInputStream to compress the data right when reading:

private static byte[] compressIntoByteArray(InputStream inputStream) throws IOException {
    try(DeflaterInputStream deflate = new DeflaterInputStream(inputStream)) {
        return deflate.readAllBytes();
    }
}

When you have the array containing the compressed data, you can use new InflaterInputStream(new ByteArrayInputStream(array)) to get an InputStream decompressing on the fly, replicating the data of the original input stream.