Question

I’m working in Hadoop, and I need to provide a comparator to sort objects as raw network order byte arrays. This is easy for me to do with integers — I just compare each byte in order. I also need to do this for floats. I think, but I can’t find a reference, that the IEEE 754 format for floats used by Java can be sorted by just comparing each byte as a signed 8 bit value.

Can anyone confirm or refute this?

Edit: the representation is IEEE 754 32 bit floating point. I actually have a (larger) byte buffer and an offset and length within that buffer. I found some utility methods already there that make it easy to turn this into a float, so I guess this question is moot. I’m still curious if anyone knows the answer.

Answer 1

Positive floats have the same ordering as their bit representations viewed as 2s complement integers. Negative floats do not.

For example, the bit representation of -2.0f is 0xc0000000 and -1.0f is 0xbf800000. If you try to use a comparison on the representations, you get -2.0f > -1.0f, which is incorrect.

There’s also the issue of NaNs (which compare unordered against all floating point data, whereas the representations do not), but you may not care about them.