How to perform round to even with floating point numbers

 

In regards to IEEE-754 single precision floating point, how do you perform round to nearest, where ties round to the nearest even digit in the required position (the default and by far the most common mode)?

Basically I have the guard bit, round bit, and sticky bit. So if we form those into a vector and call it GRS, then the following rules apply:

  • If G = 0 , round down (do nothing)
  • If G = 1 , and RS == 10 or RS == 01 , round up (add one to mantissa)
  • if GSR = 111 , round to even

    • So I am not sure how to perform the round to nearest. Any help is greatly appreciated.

    Just to make sure we're on the same page, G is the most significant bit of the three, R comes next and S can be thought of as the least significant bit because its value partially represents the even less significant bits that have been truncated in the calculations. These three bits are only used while doing calculations and aren't stored in the floating-point variable before or after the calculations.

    This is what you should do in order to round the result to the nearest even number using G , R and S :

    GRS - Action
    0xx - round down = do nothing (x means any bit value, 0 or 1)
    100 - this is a tie : round up if the mantissa's bit just before G is 1, else round down=do nothing
    101 - round up
    110 - round up
    111 - round up

    Rounding up is done by adding 1 to the mantissa in the mantissa's least significant bit position just before G . If the mantissa overflows (its 23 least significant bits that you will store become zeroes), you have to add 1 to the exponent. If the exponent overflows, you set the number to +infinity or -infinity depending on the number's sign.

    In the case of a tie, you add 1 to the mantissa if the mantissa is odd and you add nothing if it's even. That's what makes the result rounded to the nearest even value.

    Just wanted to add that S bit is not just a bit following GR bits. If there are bits available after GRS bits, it actually is a logical OR of those, including S bit.
    In other words, if there is any bit following GR bits that is 1 then the S bit value will be 1 .

    链接地址: http://www.djcxy.com/p/85630.html

    上一篇: C或C ++中的十进制算术？

    下一篇: 如何使用浮点数执行循环