CODEWITHSUNDEEP
javaporting
Corey Ogburn
Corey Ogburn

Reputation: 24717

Converting long[64] to byte[512] in Java?

I'm porting a process over to Java. There's already working versions in C# and C++.

I have a section in C# that I do Marshal.Copy(...) to convert 64 ulongs to 512 bytes and that line in C++ I use memmove(...) to do the same thing. What is available in Java to achieve the same result? I need the same binary information in the same order just as bytes instead of longs.

Edit:

The reason I'm porting to Java is to take advantage of the portability that Java naturally has. I would not like to use native code.

Another thing. Since Java doesn't contain unsigned values, then I need to change what I'm requesting by just a little. I would like to attain the 8 unsigned byte values from each of the 64 longs (ulongs in C# and C++) so that I can use those values at indices in arrays later. This needs to happen thousands of times so the fastest way would be the best way.

Upvotes: 17

Views: 3929

Answers (1)

trashgod
trashgod

Reputation: 205785

ByteBuffer works well for this: just put in 64 long values and get a byte[] out using the array() method. The ByteOrder class can handle endian issues effectively. For example, incorporating the approach suggested in a comment by wierob:

private static byte[] xform(long[] la, ByteOrder order) {
    ByteBuffer bb = ByteBuffer.allocate(la.length * 8);
    bb.order(order);
    bb.asLongBuffer().put(la);
    return bb.array();
}

Addendum: The resulting byte[] components are signed, 8-bit values, but Java arrays require nonnegative integer index values. Casting a byte to an int will result in sign extension, but masking the higher order bits will give the unsigned value of byte b:

int i = (int) b & 0xFF;

This answer elaborates on the applicable operator precedence rules. This related answer demonstrates a similar approach for double values.

Upvotes: 21

Related Questions