Odd result when bit shifting in Java

Question

I'm trying to isolate two bytes that are next to each other add them, but there seems to be an extra bit that sometimes shows up and I can't figure out how to get rid of it. It's throwing off the answer. The code is:

(acc & 0x00000000ff000000L) + ((acc << 8) & 0x00000000ff000000L);

and I'm getting results such as

0x0000000147000000

when it should be

0x0000000047000000

How can I get rid of the 1?

Edit: acc is a long. I'm try to add the 5th and 6th byte and then that value will go into a new long in the 5th byte position.

Answer 1

You need to mask the bits you want at the end, because the addition may carry a bit:

((acc & 0x00000000ff000000L) + ((acc << 8) & 0x00000000ff000000L)) & 0x00000000ff000000L;

I think this might be clearer if you broke it down a little:

acc&=0x00000000FFFF000000L; // isolate bytes 5 and 4
acc+=(acc<<8);              // add the two bytes (we'll strip bytes 6 & 4 next)
acc&=0x00000000FF00000000L; // reduce to byte 5 only

which happens to be one less bitwise opperation too.

Answer 2

If I understand correctly, you want to get the value of the 5th and 6th byte, add them together, and store them in a new long that contains just that sum in the 5th byte. This would be done like this:

long 5thByte = acc & 0xff00000000 >>> 32;
long 6thByte = acc & 0xff0000000000 >>> 40;
long sum = 5thByte + 6thByte;
long longWithNewByte = sum << 32;

This will of course carryover to the 6th byte if the sum is higher than 255. To get rid of that carryover, you can use another mask.

long longWithNewByte &= 0xff00000000;