这个浮点数加法算法究竟哪里不对？

[rcocco]

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

1. 指数大的减小的得到移位量10000001000-01111111100=1100，即 12 。
2. 把较小值的尾数右移 12 位得到

   0.0000000000011001100110011001100110011001100110011001 100
   

3. 尾数相加

   1.1111010000001100110011001100110011001100110011001101 000 +
   0.0000000000011001100110011001100110011001100110011001 100
   ==========================================================
   1.1111010000100110011001100110011001100110011001100110 100
   

4. 隐含位为 1，不需要再调整，得到

   0 10000001000 1.1111010000100110011001100110011001100110011001100110 100
   

5. 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));

[no1xsyzy]

猜想是右移 12 位后最后 1001 101 吧？
那样就进了 1

[rcocco]

@no1xsyzy 对，我忘了 sticky bit 置 1 后哪怕再移位也是不变的。