IEEE中逐渐下溢和非规格化的数字 [英] Gradual underflow and denormalized numbers in IEEE

问题描述

我正在阅读关于浮点表示和下溢/溢出的问题，并且我想到了一些有趣的东西 - 逐渐下溢。根据我的理解，渐进式下溢意味着例如减法运算x-y的结果非常小，以至于可以将其清除为0，但浮点系统产生的数量比UFL小。到处都是我读的，它是通过丢失一些准确的，这意味着一些尾数的尾数进行指数，所以我们可以有一个更小的指数？ 解决方案

>有效的答案是肯定的 - 尾数的位到指数。这些被称为次正常（又名非正常）数字。例如，在IEEE双精度中，正常数的两个指数的最小次幂（具有全部53位精度的数）是2·1022·/ sup>。但是可以有效地表示2到2〜1074的幂，如非规范化的有效位中前1位的位置所规定的。所以指数2〜1023有52位精度，2〜1024有51位精度，...，2〜-1074有1一点点的精度。

（看我的文章两台计算机的外观看起来更像是一台计算机。

I was reading about floating-point representation and underflow/overflow and I ecnountered something interesting - gradual underflow. As I understand gradual underflow means that the result of, for example substraction x-y is so small that it could be flushed to 0 but floating-point system produces number that is smaller then UFL. Everywhere I read that it is made by losing some precission, it means that some bits of mantissa goes to exponent so we can have smaller exponent?

解决方案

Effectively the answer is yes -- the bits of the mantissa go to the exponent. These are called subnormal (AKA denormal) numbers. For example, in IEEE double-precision, the smallest power of two exponent for a normal number -- a number with a full 53 bits of precision -- is 2^-1022. But powers of two up to 2^-1074 can effectively be represented, as dictated by the location of the first 1 bit in the unnormalized significand. So exponent 2^-1023 has 52 bits of precision, 2^-1024 has 51 bits of precision, ... , 2^-1074 has 1 bit of precision.

(See my article What Powers of Two Look Like Inside a Computer to visualize this better.)

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