问题
I'm trying to wrap my head around this floating point representation of binary numbers, but I couldn't find, no matter where I looked, a good answer to the question.
Why is the exponent biased?
What's wrong with the good old reliable two's complement method?
I tried to look at the Wikipedia's article regarding the topic, but all it says is: "the usual representation for signed values, would make comparison harder."
回答1:
The IEEE 754 encodings have a convenient property that an order comparison can be performed between two positive non-NaN numbers by simply comparing the corresponding bit strings lexicographically, or equivalently, by interpreting those bit strings as unsigned integers and comparing those integers. This works across the entire floating-point range from +0.0 to +Infinity (and then it's a simple matter to extend the comparison to take sign into account). Thus for example in IEEE 754 binary 64 format, 1.1 is encoded as the bit string (msb first)
0011111111110001100110011001100110011001100110011001100110011010
while 0.01 is encoded as the bit string
0011111110000100011110101110000101000111101011100001010001111011
which occurs lexicographically before the bit string for 1.1.
For this to work, numbers with smaller exponents need to compare before numbers with larger exponents. A biased exponent makes that work, while an exponent represented in two's complement would make the comparison more involved. I believe this is what the Wikipedia comment applies to.
Another observation is that with the chosen encoding, the floating-point number +0.0 is encoded as a bit string consisting entirely of zeros.
回答2:
I do not recall the specifics, but there was some desire for the highest exponent to be slightly farther from zero than the least normal exponent. This increases the number of values x for which both x and its reciprocal are approximately representable. For example, with IEEE-754 64-bit binary floating-point, the normal exponent range is -1022 to 1023. This makes the largest finite representable value just under 21024, so the interval for which x and its reciprocal are both approximately representable is almost 2-1024 to almost 21024. (Numbers at the very low end of this interval are subnormal, so some precision is being lost, but they are still representable.)
With a two’s complement representation, the exponent values would range from -1024 to 1023, and we have to reserve two of them to handle zeros, subnormals, infinities, and NaNs. That leaves a range of -1023 to 1022. With this, the interval for x such that both x and its reciprocal are approximately representable is almost 2-1023 to 21023. Thus, the biased arrangement provides a greater useful range of values.
来源：https://stackoverflow.com/questions/19864749/why-do-we-bias-the-exponent-of-a-floating-point-number