javascript 解决精度丢失问题

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  • Post category:java

加法:

var add = function(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
};

减法:

var subtraction = function(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));
    //动态控制精度长度
    n = (r1 >= r2) ? r1 : r2;
    return ((arg1 * m - arg2 * m) / m).toFixed(n);
};

 


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