byte[] length from LargeInteger.bitLength()

Question

I'm trying to convert a LargeInteger into a byte[] of unknown length using bitLength() with

static byte[] encodeint(LargeInteger y) {
    //byte[] in = y.toByteArray();
    byte[] in = new byte[(int)Math.ceil((double)y.bitLength() / 8.0)];
    y.toByteArray(in, 0);
    //
    byte[] out = new byte[in.length];
    for (int i=0;i<in.length;i++) {
        out[i] = in[in.length-1-i];
    }
    return out;
}

but the executor returns

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0

which points to y.toByteArray(in, 0);.

How can the length of in be properly set?

(The commented code is leftover from converted BigInteger code.)

BTW, it's faster, more compact, and likely more robust to compute the number of bytes as `(y.bitLength() + 7) / 8`. — Hot Licks, Jan 22 '14 at 18:25
@HotLicks Hmm, it didn't like that. Are you sure it's correct? Thank you very much in advance! — , Jan 22 '14 at 18:29

score 2 · Accepted Answer · answered Jan 22 '14 at 18:04

2

The javadoc for toByteArray tells you

java.lang.IndexOutOfBoundsException - if bytes.length < (bitLength() >> 3) + 1

Thus in should be >= (bitLength() >> 3) + 1

What you have done is nearly the same except you have not added the 1.

So (int)Math.ceil((double)y.bitLength() / 8.0) -1

but easier to use the documented version y.(bitLength() >> 3) + 1

answered Jan 22 '14 at 18:04

mmmmmm

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Thank you Mark! I was thinking it was due to the remainder. I think for this code (ed25519), it should be as precise as possible, but I don't know. Would you mind adding a snippet to add `1` if there's a remainder? – Jan 22 '14 at 18:07
@Gracchus = the documented version is a precise as possible as if any less you will get the exception – mmmmmm Jan 22 '14 at 18:08

byte[] length from LargeInteger.bitLength()

1 Answers1