Question

我在Javascript中有一个双倍的值，例如，1.0883076389305e-311。我想以下面的形式表达它，使用'bc'实用程序来计算扩展/更高精度/比例形式：

$ bc
scale=400
1.0883076389305000*10^-311
.0000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000000000000000000000000000000000\
00000000000000000000000000000000000000010883076389305000000000000000\
0000000000000000000000000000000000000000000000000000000000000

我需要一个Javascript bigint库或代码来生成与字符串相同的输出，并使用数字的扩展/更高精度格式。

谢谢！

Answer 1

这很糟糕，但适用于我能想到的每个测试用例：

Number.prototype.toFullFixed = function() {
    var s = Math.abs(this).toExponential();
    var a = s.split('e');
    var f = a[0].replace('.', '');
    var d = f.length;
    var e = parseInt(a[1], 10);
    var n = Math.abs(e);

    if (e >= 0) {
        n = n - d + 1;
    }

    var z = '';
    for (var i = 0; i < n; ++i) {
        z += '0';
    }

    if (e <= 0) {
        f = z + f;
        f = f.substring(0, 1) + '.' + f.substring(1);
    } else {
        f = f + z;
        if (n < 0) {
            f = f.substring(0, e + 1) + '.' + f.substring(e + 1);
        }
    }

    if (this < 0) {
        f = '-' + f;
    }

    return f;
};

如果您发现某个号码无法正确解析，即n !== parseFloat(n.toFullFixed())，请告诉我它是什么！

Answer 2

//只要处理数字字符串而不是数字就可以了使用字符串方法将指数幅度和精度转换为零

function longPrecision(n, p){
    if(typeof n== 'string'){
        n= n.replace('*10', '').replace('^', 'e');
    }
    p= p || 0;
    var data= String(n), mag, str, sign, z= '';
    if(!/[eE]/.test(data)){
        return data;
        if(data.indexOf('.')== -1 && data.length<p) data+= '.0';
        while(data.length<p) data+= '0';
        return data;
    }
    data= data.split(/[eE]/);
    str= data[0];
    sign= str.charAt(0)== "-"? "-": "";
    str= str.replace(/(^[+-])|\./, "");
    mag= Number(data[1])+ 1;
    if(mag < 0){
        z= sign + "0.";
        while(mag++) z += "0";
        str= z+str;
        while(str.length<p) str+= '0';
        return str;
    }
    mag -= str.length;
    str= sign+str;
    while(mag--) z += "0";
    str += z;
    if(str.indexOf('.')== -1 && str.length<p) str+= '.0';
    while(str.length<p) str+= '0';
    return str;
}


var n='1.0883076389305000*10^-311';
longPrecision(n, 400);


/*  returned value: (String)
0.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001088307638930500000000000000000000000000000000000000000000000000000000000000000000000000
*/

在Javascript中以非指数/短格式转换/表示双数

2 个答案: