FE.ES-终结0.1+0.2,答到点上的那种

随便逛了一下知乎和思否的博文，大都没回答到点上。I.浮点数的二进制存储采用 IEEE 754 规范来存储浮点数：1位【正负符号】+11位【指数】+52位【有效数字】，如下图由于0.1.toString(2)=0.0001100110011001100110011001100110011001100110011001101所以0.1 = 2^-4 * [1].10011001100110011001100110011001100110011001100110100.2 = 2^-3 * [1].1001100110011001100110011001100110011001100110011010II. 到底怎么相加铁律是52位有效数字，也就是：0.1 = 2^-3 * 0.1100110011001100110011001100110011001100110011001101(0)0.2 = 2^-3 * 1.1001100110011001100110011001100110011001100110011010sum = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)由于有效数字变成了53位，根据IEEE754 rounding mode 的 Round to Nearest,若x在a和b之间，选择最低有效位为零的值。a = 2^-2 * 1.0011001100110011001100110011001100110011001100110011x = 2^-2 * 1.0011001100110011001100110011001100110011001100110011(1)b = 2^-2 * 1.0011001100110011001100110011001100110011001100110100当比较0.1+0.2 和 0.3时，实际比较的是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…结果很明显啦参考资料：https://stackoverflow.com/que…