这个浮点数加法算法究竟哪里不对? - V2EX
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rcocco
V2EX    程序员

这个浮点数加法算法究竟哪里不对?

  •  
  •   rcocco 2020-11-24 20:28:32 +08:00 1743 次点击
    这是一个创建于 1789 天前的主题,其中的信息可能已经有所发展或是发生改变。

    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)); 
    2 条回复    2020-11-24 20:59:53 +08:00
    no1xsyzy
        1
    no1xsyzy  
       2020-11-24 20:44:26 +08:00
    猜想是右移 12 位后最后 1001 101 吧?
    那样就进了 1
    rcocco
        2
    rcocco  
    OP
       2020-11-24 20:59:53 +08:00
    @no1xsyzy 对,我忘了 sticky bit 置 1 后哪怕再移位也是不变的。
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