计算1000.1+0.2,把尾数前面隐含的 1 位也写出来,也加上 Guard bit 、Round bit 和 Sticky bit
1000.1 =
0 10000001000 1.1111010000001100110011001100110011001100110011001101 000
0.2 =
0 01111111100 1.1001100110011001100110011001100110011001100110011010 000
- 指数大的减小的得到移位量
10000001000-01111111100=1100,即 12 。 - 把较小值的尾数右移 12 位得到
0.0000000000011001100110011001100110011001100110011001 100 - 尾数相加
1.1111010000001100110011001100110011001100110011001101 000 + 0.0000000000011001100110011001100110011001100110011001 100 ========================================================== 1.1111010000100110011001100110011001100110011001100110 100 - 隐含位为 1,不需要再调整,得到
0 10000001000 1.1111010000100110011001100110011001100110011001100110 100 - Guard bit 、Round bit 和 Sticky bit 是
100中间值,向偶数舍入,使最低有效位为 00 10000001000 1111010000100110011001100110011001100110011001100110
为啥这样算出来的值和 JavaScript 中1000.1+0.2差 1 个 bit ? JS 中得到的值是下面这个
0 10000001000 1111010000100110011001100110011001100110011001100111
和之前算的就差 1 个 bit
0 10000001000 1111010000100110011001100110011001100110011001100110
let d = new DataView(new ArrayBuffer(8));
d.setFloat64(0, 1000.1+0.2);
console.log(d.getBigUint64(0).toString(16));