将最大整数值存储为double的问题

你不能，在实践中。 double不具有存储大数字的精度。

您可以通过添加D后缀来编译它以使编译器将其视为double而不是整数：

You can't, in practice. A double doesn't have the precision to store a number that big.
You can get it to compile by adding a "D" suffix to make the compiler treat it as a double rather than an integer:

double asa =12123435352352349879798724395792347897947323979743526D;
Console.WriteLine(   "12123435352352349879798724395792347897947323979743526");
Console.WriteLine(asa);
Console.WriteLine("{0:00000000000000000000000000000000000000000000000000000}", asa);

但它打印的内容可能不是你想要的：

But what it prints is probably not what you want:

12123435352352349879798724395792347897947323979743526
1.21234353523524E+52
12123435352352400000000000000000000000000000000000000

如何打印12123435352352349879798724395792347897947323979743526像这样......



你不能用双 - 或浮点数，十进制数，甚至长数字打印它 - 它作为一个整数超过64位！ />
如果添加对System.Numerics的引用并使用BigInteger，则可以将其用作数字：

"how to print it 12123435352352349879798724395792347897947323979743526 like this..."

You can't print it from a double - or a float, or decimal, or even a long - it exceeds 64 bits as an integer!
You can use it as a number if you add a reference to System.Numerics and use BigInteger:

double asa = 12123435352352349879798724395792347897947323979743526D;
BigInteger bi = BigInteger.Parse("12123435352352349879798724395792347897947323979743526");
Console.WriteLine("12123435352352349879798724395792347897947323979743526");
Console.WriteLine(asa);
Console.WriteLine("{0:00000000000000000000000000000000000000000000000000000}", asa);
Console.WriteLine(bi);

12123435352352349879798724395792347897947323979743526
1.21234353523524E+52
12123435352352400000000000000000000000000000000000000
12123435352352349879798724395792347897947323979743526

但请注意 - 这不是一个快速的课程！

But be aware - this is not a "quick" class!