5 ms·

So, when processing bytes, can you name an operation where the existence of an unsigned type would even make a difference? I'm still not seeing it.

• eMSF 5y ago
  Java bytes are signed. They aren't useful when working with unsigned 8-bit values (say, a scaling factor or an index in whatever binary format you are trying to parse). You need a larger integer type, and indeed that bitwise and operation to get the range of values you need.
• xxs 5y ago
  java virtually has only 2 integer types int and long, or 4 and 8 bytes one. So all operations are converted to int (or long) 1st, even when they involve byte/short, etc. Imagine you try to combine a 4bytes (from an array) into a single int. If you just bitwise OR and shift left, e.g. something like (b[0] << 24) | (b[1] << 16) | (b[2] << 8) | b[3] it won't work, you need ((b[1] & 0ff) << 16), etc. The byte values greater than 127 would have the sign bit set (and all other higher bits to make the 2complimentary form) That part is somewhat alleviated by wrapping byte arrays into ByteBuffers that would the right thing, or even better using only ByteBuffers (preferably the direct version of them when reading from input/output)