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It avoids error whenever possible. For example, if you are working with pure fractions the result will be exactly representable and you'll end with 1/3 + 1/3 +
by Maken 11y ago
It avoids error whenever possible. For example, if you are working with pure fractions the result will be exactly representable and you'll end with 1/3 + 1/3 + 1/3 = 1 instead of some approximation.
- Someone 11y agoThat would be nice, but 1/3 isn't exactly representable in this representation. Or are you suggesting that, when computing 1/3, this number format picks the best representation for a given number of bits, keeping the 'exact' bit, for 1/3, and then magically decides that, say, 85/256 and 'exact' must be 1/3, and not 85/256, or whatever other expression that could be computed exactly gets close to it?