递归-PrintDigit

static void PrintDigit(double n,int len,int count=-1)
{
    if (n <0) { n = System.Math.Abs(n); Console.Write("-"); }
    int a = (int)(n * (Math.Pow(10,count)) % 10);
    Console.Write(a);
    if (count == 0) Console.Write(".");
    count++;
    if (count > len) return;
    else
         PrintDigit(n, len,count);
}

这位大佬的讲解真好。