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javabinarymasking

Java four bytes to binary 'bitmask'

I'm trying to take four bytes from a binary file and turn them into a bit mask that represents a zero for data and a 1 for metadata.

I'm assuming I need to convert them into an int by or'ing them together and then bit shift through that to determine if a bit position is set or not.

I can't work out how to combine all four together to get a 32-bit mask that I can step though.

int mask = (((int)hbyte) & 0xff) << 8 | (((int)lbyte) & 0xff);

That would combine two together (post by Laurence Gonsalves from another post) but I can't figure out the syntax to get four together.

Or would this be case for an Enumset, which I'll admit I don't fully understand.

Any advice would be appreciated.

Thanks.

Tony

**Just to add another quick note to this, (I'm not sure if that's generally allowed so I apologise in advance if it's not), what is the best way of going through my newly created int and checking if a bit is set or not?

Currently I'm usingif (((comboOfBytes >> bitmaskBit) & 1) == 0) but I don't know if that's the most elegant way of doing the check that I need.

Also I'm still not entirely sure if I understand how shifting actually allows all 32 bits to be checked!

Thanks again

Solution

  • You can use a DataInputStream to read 4 bytes as an int from your file.

    Alternately, if you read bytes a, b, c, d:

    int comboOfBytes = ((a & 0xff) << 24) | ((b & 0xff) << 16) | ((c & 0xff) << 8) | (d & 0xff);

    The (& 0xff) is necessary to prevent sign bit extension from clobbering higher bytes.