Converting C# code to Java

Question

So I tried to convert a piece of C# code to Java but I do not get same output when I use the converted code.

I already searched google and couldn't really find an answer. I' mostly interested in the Java equivalent of the TransformBlock method as I think that it is the one which causes problems.

C# code:

public class Sha256
{
    public Sha256()
    {
        sha = new SHA256Managed();
        sha.Initialize();
    }

    public void Process(byte[] data, int length)
    {
        sha.TransformBlock(data, 0, length, data, 0);
    }

    public void Process(uint data)
    {
        var bytes = BitConverter.GetBytes(data);

        sha.TransformBlock(bytes, 0, 4, bytes, 0);
    }

    public void Process(string data)
    {
        var bytes = Encoding.UTF8.GetBytes(data);

        sha.TransformBlock(bytes, 0, bytes.Length, bytes, 0);
    }

    public void Finish(byte[] data)
    {
        sha.TransformFinalBlock(data, 0, data.Length);

        Digest = sha.Hash;
    }

    public void Finish(byte[] data, int offset, int length)
    {
        sha.TransformFinalBlock(data, offset, length);

        Digest = sha.Hash;
    }

    SHA256 sha;
    public byte[] Digest { get; private set; }
}

Java code:

public class Sha256 {

    public byte[] digest;
    private MessageDigest sha;

    public Sha256() {
        try {
            sha = MessageDigest.getInstance("SHA-256");
        } catch (NoSuchAlgorithmException ex) {
            Logger.logException(ex);
        }
    }

    public void process(byte[] data, int length) {
        sha.update(data, 0, length);
    }

    public void process(int data) {
        byte[] bytes = ByteBuffer.allocate(Integer.BYTES).putInt(data).array();

        sha.update(bytes, 0, 4);
    }

    public void process(String data) {
        byte[] bytes = data.getBytes(StandardCharsets.UTF_8);

        sha.update(bytes, 0, bytes.length);
    }

    public void finish(byte[] data) {
        sha.update(data, 0, data.length);

        digest = sha.digest();
    }

    public void finish(byte[] data, int offset, int length) {
        sha.update(data, offset, length);

        digest = sha.digest();
    }

}

You have many different method calls in there, each one of which could change when you port to Java. In any case, I don't think we can help you , because you didn't also show your Java code. — Tim Biegeleisen
– Tim Biegeleisen, Commented Jul 18, 2018 at 11:10
"I do not get same output" -- What do you get? What do you expect? For what input? — khelwood
– khelwood, Commented Jul 18, 2018 at 11:24

Peter Lawrey · Accepted Answer · 2018-07-18 15:52:57Z

1

C# is little endian by default AFAIK. This is most likely as it was developed by Microsoft who in turn used x86/x64 machines where little endian is the default.

Java's ByteBuffer uses big endian by default. It was first developed on SPARC processors which are big endian. There is couple of ways to swap the order of the bytes but the most natural is to use ByteBuffer.order() as below.

public void process(int data) {
    byte[] bytes = ByteBuffer.allocate(Integer.BYTES)
                             .order(ByteOrder.LITTLE_ENDIAN)
                             .putInt(data)
                             .array();

    sha.update(bytes, 0, bytes.length);
}

answered Jul 18, 2018 at 15:52

Peter Lawrey

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