将"大"十六进制数字(字符串格式)转换为十进制数字(字符串格式),不带BigInteger类

Question

如何转换"大"十六进制数字(字符串格式):

EC851A69B8ACD843164E10CFF70CF9E86DC2FEE3CF6F374B43C854E3342A2F1AC3E30C741CC41E679DF6D07CE6FA3A66083EC9B8C8BF3AF05D8BDBB0AA6CB3EF8C5BAA2A5E531BA9E28592F99E0FE4F95169A6C63F635D0197E325C5EC76219B907E4EBDCD401FB1986E4E3CA661FF73E7E2B8FD9988E753B7042B2BBCA76679

到十进制数字(字符串格式):

166089946137986168535368849184301740204613753693156360462575217560130904921953976324839782808018277000296027060873747803291797869684516494894741699267674246881622658654267131250470956587908385447044319923040838072975636163137212887824248575510341104029461758594855159174329892125993844566497176102668262139513

不使用BigIntegerClass(因为我的应用程序应该支持没有.NET Framework 4的机器)？

Answer 1

这是一个快速而肮脏的实现,可以处理任意大的数字.这种实现的目的是简单,而不是性能; 因此,如果要在生产场景中使用它,应该大大优化它.

编辑:根据DanByström实施的反十进制到十六进制转换进一步简化:

static string HexToDecimal(string hex)
{
    List<int> dec = new List<int> { 0 };   // decimal result

    foreach (char c in hex)
    {
        int carry = Convert.ToInt32(c.ToString(), 16);   
            // initially holds decimal value of current hex digit;
            // subsequently holds carry-over for multiplication

        for (int i = 0; i < dec.Count; ++i)
        {
            int val = dec[i] * 16 + carry;
            dec[i] = val % 10;
            carry = val / 10;
        }

        while (carry > 0)
        {
            dec.Add(carry % 10);
            carry /= 10;
        }
    }

    var chars = dec.Select(d => (char)('0' + d));
    var cArr = chars.Reverse().ToArray();
    return new string(cArr);
}