js double 精度损失 bugs All In One

js double 精度损失 bugs All In One

const arr = [
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01
];

// [0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01]

arr.reduce((acc, i) => acc += i);
// 0.09999999999999999

arr = [
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01
]
(10) [0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01]

arr.reduce((acc, i) => acc += i);
0.09999999999999999

sum = 0;
0
sum += 0.01;
0.01
sum += 0.01;
0.02
sum += 0.01;
0.03
sum += 0.01;
0.04
sum += 0.01;
0.05
sum += 0.01;
0.060000000000000005
sum += 0.01;
0.07
sum += 0.01;
0.08
sum += 0.01;
0.09
sum += 0.01;
0.09999999999999999

// 保留两位精度 ？

解决方案

1. string 大数相加 / 大数相乘

arr = [
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01
];
(10) [0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01]

arr.reduce((acc, i) => acc += Number.parseFloat(i).toFixed(2));
"0.010.010.010.010.010.010.010.010.010.01"

arr.reduce((acc, i) => acc += parseFloat(i.toFixed(2)));
0.09999999999999999

arr.reduce((acc, i) => acc += i.toFixed(2));
"0.010.010.010.010.010.010.010.010.010.01"

arr.map(i => i.toFixed(2));
(10) ["0.01", "0.01", "0.01", "0.01", "0.01", "0.01", "0.01", "0.01", "0.01", "0.01"]

// string 大数相加, ??? 位运算

2. 小数转整数

arr = [
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01, 0.01, 0.01,
  0.01
];
(10) [0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01, 0.01]
arr.map(i => i * 100).reduce((acc, i) => acc += i) / 100;
0.1

红包算法

refs

最简单的方法实现微信红包的随机算法

https://www.cnblogs.com/xgqfrms/p/13688375.html

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