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    -1 I'm not asking about the sizes of the fields in numbers of bits, just the actual order of them. The size of the single is not relevant to the question at all, and instead I'm asking why we are fitting the sign bit with the seven high bits of the exponent into a byte. Reordering them such that all eight bits of the exponent be in one byte, and the sign moved into one of the mantissa bytes, doesn't require changing anything about either the precision or range of the single Commented Feb 28 at 14:26
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    @petroleus your question about bit ordering to match byte boundaries is only relevant with certain number of bits, so the number of bits is very much relevant. Second, we give answers to questions to help everyone, not as a personal answering service to you. Commented Feb 28 at 18:45
  • @petroleus, I think the meat of gnasher's answer is in the last paragraph. The first two paragraphs are just providing a mathematical basis for the conclusion reached in the last paragraph: aligning to 8-bits has no advantage, nobody is complaining about it, and it would only matter on 8-bit hardware that nobody uses in the real world. Assuming I understood the answer correctly, maybe an edit to this answer to clarify or highlight this conclusion would help? Commented Feb 28 at 20:36
  • If you had an exponent field with nine bits, eight bit alignment would mean you have 6 mantissa bits before and 16 after the exponent field. Which would be utterly stupid. Commented Sep 2 at 18:19