这个浮点数加法算法究竟哪里不对？

資深大佬 : rcocco 1

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 0
```
0 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));

大佬有話說 (2)