问题
I have this binary representation of 0.1:
0.00011001100110011001100110011001100110011001100110011001100110
I need to round it to the nearest even to be able to store it in the double precision floating point. I can't seem to understand how to do that. Most tutorials talk about guard, round and sticky bits - where are they in this representation?
Also I've found the following explanation:
Let’s see what 0.1 looks like in double-precision. First, let’s write it in binary, truncated to 57 significant bits:
0.000110011001100110011001100110011001100110011001100110011001…
Bits 54 and beyond total to greater than half the value of bit position 53, so this rounds up to
0.0001100110011001100110011001100110011001100110011001101
This one doesn't talk about GRS bits, why? Aren't they always required?
回答1:
The text you quote is from my article Why 0.1 Does Not Exist In Floating-Point . In that article I am showing how to do the conversion by hand, and the "GRS" bits are an IEEE implementation detail. Even if you are using a computer to do the conversion, you don't have to use IEEE arithmetic (and you shouldn't if you want to do it correctly ), so the GRS bits won't come into play there either. In any case, the GRS bits apply to calculations, not really to the conceptual idea of conversion.
来源：https://stackoverflow.com/questions/37111324/how-do-i-round-this-binary-number-to-the-nearest-even