问题
I am learning about floating point formats (IEEE). In the single precision floating point format ,it is mentioned that the mantissa has 24 bits and so it has 6 1/2 decimal digits of precision (as the per the book "understanding the machine") , and 7.22 decimal digits of precision.
I don't understand how the decimal digits of precision is calculated. Can somebody please enlighten me ?
回答1:
With 24 bits, assuming one bit is reserved for the sign, then the largest decimal number you can represent is 2^23-1=8388607. That is, you can get 6 digits and sometimes a 7th. This is often expressed as "6 1/2 digits". If the 24 bits are representing an unsigned number, then the maximum value you can store is 2^24-1=16,777,215, or 7 and a fraction digits.
When someone quotes you a number with explicit decimal places like 7.22 decimal digits, what they're doing is taking the log (base 10) of the maximum value. So log(16777115)=7.22.
In general, the number of decimal digits you'll get from a given number of bits is:
d=log[base 10](2^b)
where b is the number of bits and d is the number of decimal digits. Then:
d=b * log(2)
d~=b * .3010
So 24 bits gives 24 * .3010 = 7.224
来源：https://stackoverflow.com/questions/10484332/how-to-calculate-decimal-digits-of-precision-based-on-the-number-of-bits