Most bitshift solutions I have seen for converting an int to a byte array go like this:
return new byte [] {
(byte) ((i >> 24) & 0xFF),
(byte) ((i >> 16) & 0xFF),
(byte) ((i >> 8) & 0xFF),
(byte) (i & 0xFF);
}
Why the & 0xFF??
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Ancat Dubher
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3That's bit-wise-and to turn the rest of the bits to 0. – Bhesh Gurung Feb 28 '14 at 16:42
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2Actually, it's probably not necessary if you will be casting to `(byte)`, since the cast does an implicit `& 0xFF` operation. – Hot Licks Feb 28 '14 at 16:45
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possible duplicate of [How Does The Bitwise & (AND) Work In Java?](http://stackoverflow.com/questions/17256644/how-does-the-bitwise-and-work-in-java) – Bhesh Gurung Feb 28 '14 at 16:47
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1possible duplicate : http://stackoverflow.com/questions/19061544/bitwise-anding-with-0xff-is-important – Graham Griffiths Feb 28 '14 at 16:47
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This will only generate the correct value if you want big endian... Shorter way to write the same code: `ByteBuffer.allocate(4).putInt(i).array()` – fge Feb 28 '14 at 16:59
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@GrahamGriffiths: Thanks, that was illuminating. – Ancat Dubher Feb 28 '14 at 17:01
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& 0xFF is redundant and makes no difference in the given case
Evgeniy Dorofeev
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Agreed, along with some hints from the other question Graham Griffiths posted in his comments. – Ancat Dubher Feb 28 '14 at 17:03