Large double mathematical multiplication in C#

Question

I want to multiply this number:

5374711027510012111075768211110475111691021051041057653548210911210211112250867 66690120741165250567278571217510410482757487

with this number:

4956889911565576581818287977011111065876967103548749122901151091038910610511189

But when I cast the result to string I get this:

2.66418508698446E+201

Which is:

266418508698446000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 000000000000000000000000000000000000000000000000000000000000 0000000000000000000000000000000

Not exactly the precise number, those zeros represent loss of precision, am I right?

Is it possible to get the precise number (every single digit) out of this calculation using C#?

Thanks

Try looking at [BigInt](http://msdn.microsoft.com/en-us/library/system.numerics.biginteger.aspx)? — PiousVenom, Aug 12 '13 at 15:14
@hometoast: Not a duplicate of that at all. That's a language-agnostic question about *implementing* a big integer data type, not about using one in a C#-specific Framework. — jason, Aug 12 '13 at 15:17
@Jason That may be what the question intended, but that's not what the answers are, and so its answers most certainly answer this question. — Servy, Aug 12 '13 at 15:26

jason · Accepted Answer · 2013-08-12T15:22:30.773

Yes. Use BigInteger. It's designed for this purpose. The numbers you are using won't fit in the primitive integral and can't even be represented precisely in the floating-point types¹.

BigInteger m = BigInteger.Parse("374711027510012111075768211110475111691021051041057653548210911210211112250867 66690120741165250567278571217510410482757487");
BigInteger n = BigInteger.Parse("4956889911565576581818287977011111065876967103548749122901151091038910610511189");
var product = m * n;
Console.WriteLine(proudct);

¹: Single-precision floating point can represent all integers between -2^24 and 2^24 exactly because it has a 23-bit explicit plus one implicit bit mantissa; after that it loses precision. As

2^24 = (2^10)^2.4 ~ (10^3)^2.4 ~ 10^7

we lose precision for some integers after approximately seven digits.

Similarly, double-precision floating point can represent all integers between -2^53 and 2^53 exactly because it has a 52-bit explicit plus one implicit bit mantissa; after that it loses precision. As

2^53 = (2^10)^5.3 ~ (10^3)^5.3 ~ 10^16

we lose precision for some integers after approximately sixteen digits.

score 0 · Answer 2 · answered Aug 12 '13 at 15:22

0

A double has a limited maximum precision (number of significant digits) of 15. Your numbers have too many digits and thus cannot be stored as double without loss of precision.

answered Aug 12 '13 at 15:22

Sebastian Negraszus

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score 0 · Answer 3 · answered Aug 12 '13 at 15:28

If you are using .NET 4.5, make use of system.Numerics assembly. http://msdn.microsoft.com/en-us/library/system.numerics.aspx

var num1 = BigInteger.Parse("5374711027510012111075768211110475111" + "69102105104105765354821091121021111225086766690120741165250567278571217510410482757487");

var num2 = BigInteger.Parse("4956889911565576581818287977011111065876967103548749122901151091038910610511189");

Console.WriteLine(num1 + num2);

Large double mathematical multiplication in C#

3 Answers3