5 ms·

> ...If you shift that range down so that 0 is in the middle of the range of values instead of the minimum... Not a downvoter, but: your concept of "shifting t

> ...If you shift that range down so that 0 is in the middle of the range of values instead of the minimum... Not a downvoter, but: your concept of "shifting the range" is also misleading. In the source domain of 16-bit numbers, [0...65535] can be split into two sets: [0...32767] [32768...65535] The first set of numbers maps to [0...32767] in 2's complement. But the second interval maps to [-32768...-1]. So it's not just a "shift" of [0...65535] onto another range. There's a discontinuous jump going from 32767 to 32768 (or -1 to 0 if converting the other direction). And actually, we don't know if the processor used 2's complement or 1's complement -- if it was 1's complement, they would have a signed 0! I think they'd have to say "remapping" the range? On the whole, I think OP did about as well as you're going to do, given the audience.

joshAg 4y ago
The range is of values, not their representation in bits, which can be mapped in any order. You could specify that the bit representation for 0 was 0x1234 and the bit representation for 1 was 0x1134 and proceeded accordingly and the range of values for those 16 bits could still independently be [-32768, 32767] or [0, 65535] or [65536, 131071] if you wanted. We know the signed int they're talking about can't be the standard 1's complement because its stated range of values is [-32768, 32767]. If the representation were 1's complement the range would be [-32767, 32767] to accommodate the -0. It could be some modified form of 1's complement where they redefine the -0 to actually be -32768, but that's not 1's complement anymore.
dragonwriter 4y ago
> And actually, we don't know if the processor used 2's complement or 1's complement -- if it was 1's complement, they would have a signed 0! We can infer it used two's complement, and absolutely rule out one's complement or any signed-zero system, because the range is [-2^15,2^15) and with a signed zero you can't represent every integer in that range in 16-bits, you have one too many unique numbers.

joshAg 4y ago

The range is of values, not their representation in bits, which can be mapped in any order. You could specify that the bit representation for 0 was 0x1234 and the bit representation for 1 was 0x1134 and proceeded accordingly and the range of values for those 16 bits could still independently be [-32768, 32767] or [0, 65535] or [65536, 131071] if you wanted. We know the signed int they're talking about can't be the standard 1's complement because its stated range of values is [-32768, 32767]. If the representation were 1's complement the range would be [-32767, 32767] to accommodate the -0. It could be some modified form of 1's complement where they redefine the -0 to actually be -32768, but that's not 1's complement anymore.

dragonwriter 4y ago

> And actually, we don't know if the processor used 2's complement or 1's complement -- if it was 1's complement, they would have a signed 0! We can infer it used two's complement, and absolutely rule out one's complement or any signed-zero system, because the range is [-2^15,2^15) and with a signed zero you can't represent every integer in that range in 16-bits, you have one too many unique numbers.