你如何处理大于 UInt64 (C#) 的数字

Question

在 C# 中，如何存储和计算显着超过 UInt64 的最大值 (18,446,744,073,709,551,615) 的数字？

Answer 1

Can you use the .NET 4.0 beta? If so, you can use BigInteger .

Otherwise, if you're sticking within 28 digits, you can use decimal - but be aware that obviously that's going to perform decimal arithmetic, so you may need to round at various places to compensate.

Answer 2

By using a BigInteger class; there's one in the the J# libraries (definitely accessible from C#), another in F# (need to test this one), and there are freestanding implementations such as this one in pure C#.

Answer 3

What is it that you wish to use these numbers for? If you are doing calculations with really big numbers, do you still need the accuracy down to the last digit? If not, you should consider using floating point values instead. They can be huge, the max value for the double type is 1.79769313486231570E+308, (in case you are not used to scientific notation it means 1.79769313486231570 multiplied by 10000000...0000 - 308 zeros).

That should be large enough for most applications

Answer 4

BigInteger represents an arbitrarily large signed integer.

using System.Numerics;

var a = BigInteger.Parse("91389681247993671255432112000000");
var b = new BigInteger(1790322312);
var c = a * b;

Answer 5

Decimal has greater range.

There is support for bigInteger in .NET 4.0 but that is still not out of beta.

Answer 6

There are several libraries for computing with big integers, most of the cryptography libraries also offer a class for that. See this for a free library.

Answer 7

Also, do check that you truly need a variable with greater capacity than Int64 and aren't falling foul of C#'s integer arithmetic.

For example, this code will yield an overflow error:

Int64 myAnswer = 20000*1024*1024;

At first glance that might seem to be because the result is too large for an Int64 variable, but actually it's because each of the numbers on the right side of the formula are implicitly typed as Int32 so the temporary memory space reserved for the result of the calculation will be Int32 size, and that's what causes the overflow.

The result will actually easily fit into an Int64 , it just needs one of the values on the right to be cast to Int64 :

Int64 myAnswer = (Int64)20000*1024*1024;

This is discussed in more detail in this answer .

(I appreciate this doesn't apply in the OP's case, but it was just this sort of issue that brought me here!)

Answer 8

You can use decimal . It is greater than Int64.

It has 28-29 significant digits.