转(JS中浮点运算精度错误BUG解决方案)

JS中的浮点运算有时会出现精度错误的BUG，比如：0.1 + 0.2 = 0.30000000000000004

解决方案如下：

// 除法

function accDiv(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 accMul((r1 / r2), Math.pow(10, t2 - t1));
 }

 

// 乘法

function accMul(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)
}

 

// 加法

function accAdd(arg1, arg2) {   
    var r1, r2, m, c;
    try { r1 = arg1.toString().split(".")[1].length } catch (e) { r1 = 0 }
    try { r2 = arg2.toString().split(".")[1].length } catch (e) { r2 = 0 }  
    c = Math.abs(r1 - r2);   
    m = Math.pow(10, Math.max(r1, r2))   
    if (c > 0) {   
        var cm = Math.pow(10, c);   
        if (r1 > r2) {   
            arg1 = Number(arg1.toString().replace(".", ""));   
            arg2 = Number(arg2.toString().replace(".", "")) * cm;   
        }   
        else {   
            arg1 = Number(arg1.toString().replace(".", "")) * cm;   
            arg2 = Number(arg2.toString().replace(".", ""));   
        }   
    }   
    else {   
        arg1 = Number(arg1.toString().replace(".", ""));   
        arg2 = Number(arg2.toString().replace(".", ""));   
    }   
    return accDiv((arg1 + arg2),m); 
}