随意逛了一下知乎和思否的博文,多数没回复到点上。
I.浮点数的二进制存储
采纳 IEEE 754 范例来存储浮点数:1
位【正负标记】+11
位【指数】+52
位【有用数字】,如下图
因为0.1.toString(2)
=0.0001100110011001100110011001100110011001100110011001101
所以
0.1 = 2^-4 * [1].1001100110011001100110011001100110011001100110011010
0.2 = 2^-3 * [1].1001100110011001100110011001100110011001100110011010
II. 究竟怎样相加
铁律是52位有用数字,也就是:
0.1 = 2^-3 * 0.1100110011001100110011001100110011001100110011001101(0)
0.2 = 2^-3 * 1.1001100110011001100110011001100110011001100110011010
sum = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)
因为有用数字变成了53位,依据IEEE754 rounding mode 的 Round to Nearest,若x在a和b之间,挑选最低有用位为零的值。
a = 2^-2 * 1.0011001100110011001100110011001100110011001100110011
x = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)
b = 2^-2 * 1.0011001100110011001100110011001100110011001100110100
当比较0.1+0.2 和 0.3时,现实比较的是
0.1 + 0.2 => 0:01111111101:0011001100110011001100110011001100110011001100110[100]
0.3 => 0:01111111101:0011001100110011001100110011001100110011001100110[011]
转成10进制,看上去是:
0.1 + 0.2 => 0.300000000000000044408920985006...
0.3 => 0.299999999999999988897769753748...
效果很明显啦