1. 浮点数IEEE 754表示方法
要搞清楚float累加为什么会产生误差，必须先大致理解float在机器里怎么存储的，这里只介绍一下组成
由上图可知(摘在[2])， 浮点数由： 符号位 + 指数位 + 尾数部分， 三部分组成。由于机器中都是由二进制存储的，那么一个10进制的小数如何表示成二进制。例如: 8.25转成二进制为1000.01, 这是因为 1000.01 = 1*2^3 + 0*2^2 + 0*2^1 + 0*2^0 + 0*2^-1 + 2*2^-2 = 1000.01.
（2）float的有效位数是6-7位，这是为什么呢？因为位数部分只有23位，所以最小的精度为1*2^-23 在10^-6和10^-7之间，接近10^-7, 中也有解释
那么为什么float累加会产生误差呢，主要原因在于两个浮点数累加的过程。
2. 两个浮点数相加的过程
两浮点数X，Y进行加减运算时，必须按以下几步执行：
（1）对阶，使两数的小数点位置对齐，小的阶码向大的阶码看齐。（2）尾数求和，将对阶后的两尾数按定点加减运算规则求和(差)。（3）规格化，为增加有效数字的位数，提高运算精度，必须将求和(差)后的尾数规格化。（4）舍入，为提高精度，要考虑尾数右移时丢失的数值位。（5）判断结果，即判断结果是否溢出。关键就在与对阶这一步骤，由于float的有效位数只有7位有效数字，如果一个大数和一个小数相加时，会产生很大的误差，因为尾数得截掉好多位。例如：
123 + 0.00023456 = 1.23*10^2 + 0.000002 * 10^2 = 123.0002
那么此时就会产生0.00003456的误差，如果累加多次，则误差就会进一步加大。
那么怎么解决这种误差呢？
function KahanSum(input) var sum = 0.0 var c = 0.0 // A running compensation for lost low-order bits. for i = 1 to input.length do var y = input[i] - c // So far, so good: c is zero. var t = sum + y // Alas, sum is big, y small, so low-order digits of y are lost. c = (t - sum) - y // (t - sum) cancels the high-order part of y; subtracting y recovers negative (low part of y) sum = t // Algebraically, c should always be zero. Beware overly-aggressive optimizing compilers! next i // Next time around, the lost low part will be added to y in a fresh attempt. return sum
例子：
1.
y = 3.14159 - 0 y = input[i] - c t = 10000.0 + 3.14159 = 10003.14159 But only six digits are retained. = 10003.1 Many digits have been lost! c = (10003.1 - 10000.0) - 3.14159 This must be evaluated as written! = 3.10000 - 3.14159 The assimilated part of y recovered, vs. the original full y. = -.0415900 Trailing zeros shown because this is six-digit arithmetic.sum = 10003.1 Thus, few digits from input(i) met those of sum.
2.
y = 2.71828 - -.0415900 The shortfall from the previous stage gets included. = 2.75987 It is of a size similar to y: most digits meet. t = 10003.1 + 2.75987 But few meet the digits of sum. = 10005.85987 And the result is rounded = 10005.9 To six digits. c = (10005.9 - 10003.1) - 2.75987 This extracts whatever went in. = 2.80000 - 2.75987 In this case, too much. = .040130 But no matter, the excess would be subtracted off next time.sum = 10005.9 Exact result is 10005.85987, this is correctly rounded to 6 digits.