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JavaScript計算精度問題

一、原因

 js的Number在記憶體中儲存使用的是一種64位雙精度浮點數儲存方法。其中,1位用來表示符號,0為正,1為負;11位用來表示指數;52位用來表示尾數。他的表示格式為:s * m * ( 2 ^ e ) (s為符號位,m為尾數,e為指數)

將十進位制的小數轉換為二進位制的小數,採用“乘二取整法”,即小數部分乘以2,取整數部分,剩下的小數部分繼續乘以二,取整數部分,直到取到小數部分為0為止。線上進位制轉換請戳這裡

舉例:0.1 + 0.2 = 0.30000000000000004

0.1(10) 可轉換成二進位制:0.0001100110011001100110011001100110011001100110011001101(2) 即:

e = -4; m = 1.1001100110011001100110011001100110011001100110011010

0.2(10) 可轉換成二進位制:0.001100110011001100110011001100110011001100110011001101(2) 即:

e = -3; m = 1.1001100110011001100110011001100110011001100110011010

二者相加:

e = -3; m = 0.1100110011001100110011001100110011001100110011001101(52位) +

e = -3; m = 1.1001100110011001100110011001100110011001100110011010(52位) 結果是:

e = -3; m = 10.0110011001100110011001100110011001100110011001100111(52位) 即

e = -2; m = 1.00110011001100110011001100110011001100110011001100111(53位) 根據round to nearest, tie to even(有近取近,無近取偶)的方法取到近似值:

e = -2; m = 1.0011001100110011001100110011001100110011001100110100(52位) 可轉換成十進位制:0.30000000000000004

 

二、解決方法

加法:

add(arg1, arg2) {
var r1, r2, m;
    try {
        r1 = arg1.toString().split(".")[1].length;
    } catch (e) {
        r1 = 0;
    }
    try {
        r2 = arg2.toString().split(".")[1].length;
    } catch (e) {
        r2 = 0;
    }
    m 
= Math.pow(10, Math.max(r1, r2)); return (arg1 * m + arg2 * m) / m; // 有問題。例如2.3 * 100 = 229.99999999999997 // return (Math.round(arg1 * m) + Math.round(arg2 * m)) / m; // 方法一:四捨五入的方式 // return (this.mul(arg1, m) + this.mul(arg2, m)) / m; // 方法二:乘法自己封裝了一層 },

減法:

sub(arg1, arg2) {
var r1, r2, m, n;
    try {
        r1 = arg1.toString.split(".")[1].length;
    } catch (e) {
        r1 = 0;
    }
    try {
        r2 = arg2.toString().split(".")[1].length;
    } catch (e) {
        r2 = 0;
    }
    m = Math.pow(10, Math.max(r1, r2));
    n = (r1 >= r2) ? r1 : r2;
    return ((arg1 * m - arg2 * m) / m).toFixed(n);
},

乘法:

mul(arg1, arg2) {
var m = 0,
    s1 = arg1.toString(),
    s2 = arg2.toString();
    try {
        m += s1.split(".")[1].length;
    } catch (e) { }
    try {
        m += s2.split(".")[1].length;
    } catch (e) { }
    return Number(s1.replace(".", "")) * Number(s2.replace(".", "")) / Math.pow(10, m);
},

除法:

div(arg1, arg2) {
    var t1 = 0, t2 = 0, r1, r2;
try {
        t1 = arg1.toString().split(".")[1].length;
    } catch (e) { }
    try {
        t2 = arg2.toString().split(".")[1].length;
    } catch (e) { }
    r1 = Number(arg1.toString().replace(".", ""));
    r2 = Number(arg2.toString().replace(".", ""));
    return (r1 / r2) * Math.pow(10, t2 - t1);
},

大整數比較大小:

value.toString().length