2 ms·
Given how floating point works, I'm not sure that the floating point representation closest to 1 would ever round to 1 (which is neatly representable). That pro
by bartonfink 16y ago
Given how floating point works, I'm not sure that the floating point representation closest to 1 would ever round to 1 (which is neatly representable). That proof works because you're working with real numbers, which have no encoding restrictions.
- CastItUp 16y agoI think you are on the right track of where I am getting... Before running through all the code, x = .9999999, but at the end of all the operations, x = 1, so; if placed in a "for loop", with a condition outside of the loop, its possible that the compiler would return "false" for x = x.