javascript數字計算丟失精度怎么解決

js數字計算丟失精度問題解決方案

計算機世界里，數字的計算，所有語言都會丟失精度，所以沒有萬全之策，但在人力范圍內，盡量解決。

網上找了一部分代碼，發現是有問題的，比如:

//加法 
Number.prototype.myAdd = function(arg2) {
    var arg1 = this;
    if (isNaN(arg2)) {
        return 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
}
//減法 
Number.prototype.mySub = function(arg2) {
    var arg1 = this;
    if (isNaN(arg2)) {
        return 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 
Number.prototype.myMul = function(arg2) {
    var arg1 = this;
    if (isNaN(arg2)) {
        return 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
Number.prototype.myDiv = function(arg2) {
    var arg1 = this;
    if (isNaN(arg2)) {
        return 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) {}
    with(Math) {
        r1 = Number(arg1.toString().replace(".", ""))
        r2 = Number(arg2.toString().replace(".", ""))
        return (r1 / r2).myMul(pow(10, t2 - t1))
    }
}

在計算一些特殊的數字時，仍然有問題：

比如加法：

268.34.myDiv(0.83);//321.7505995203837

所以還要優化

我重新做了一版：

var operationNumber = function (arg1,arg2,operator) {
    var oper=['+','-','*','/'];
    // 不合法的運算
    if (isNaN(arg1)||isNaN(arg2)||oper.indexOf(operator)<0) {
        return NaN;
    }
    // 除以0
    if (operator==='/'&&Number(arg2)===0) {
        return Infinity;
    }
    // 和0相乘
    if (operator==='*'&&Number(arg2)===0) {
        return 0;
    }
    // 相等兩個數字相減
    if ((arg1===arg2||Number(arg1)===Number(arg2))&&operator==='-') {
        return 0;
    }
    var r1, r2, max,_r1,_r2;
    try { r1 = arg1.toString().split(".")[1].length } catch (e) { r1 = 0 }
    try { r2 = arg2.toString().split(".")[1].length } catch (e) { r2 = 0 }
    max = Math.max(r1, r2)
    _r1 = max-r1;
    _r2 = max-r2;
    if (_r1!==0) {
        arg1=arg1+'0'.repeat(_r1)
    }
    if (_r2!==0) {
        arg2=arg2+'0'.repeat(_r2)
    }
    arg1 = Number(arg1.toString().replace('.',''))
    arg2 = Number(arg2.toString().replace('.',''))
    var r3 = operator==='*'?(max*2):(operator==='/'?0:max);
    var newNum = eval(arg1+operator+arg2);
    if (r3!==0) {
        var nStr = newNum.toString();
        nStr = nStr.replace(/^-/,'');
        if (nStr.length<r3+1) {
            nStr = '0'.repeat(r3+1-nStr.length)+nStr;
        }
        nStr = nStr.replace(new RegExp('(\\\d{'+r3+'})$'),'.$1');
        if (newNum<0) {
            nStr = '-'+nStr;
        }
        newNum = nStr*1;
    }
    return newNum;
}
//加法 
Number.prototype.myAdd = function(arg2) {
    return operationNumber(this,arg2,'+');
}
//減法 
Number.prototype.mySub = function(arg2) {
    return operationNumber(this,arg2,'-');
}
//乘法 
Number.prototype.myMul = function(arg2) {
    return operationNumber(this,arg2,'*');
}
// 除法
Number.prototype.myDiv = function(arg2) {
    return operationNumber(this,arg2,'/');
}

以上就是怎么解決javascript數字計算丟失精度問題？的詳細內容，更多請關注億速云其它相關文章！