C# Convert.ToDouble() loses decimal points when converting string to double

Question

Let's say we have the following simple code

        string number = "93389.429999999993";
        double numberAsDouble = Convert.ToDouble(number);
        Console.WriteLine(numberAsDouble);

after that conversion numberAsDouble variable has the value 93389.43. What can i do to make this variable keep the full number as is without rounding it? I have found that Convert.ToDecimal does not behave the same way but i need to have the value as double.

-------------------small update---------------------

putting a breakpoint in line 2 of the above code shows that the numberAsDouble variable has the rounded value 93389.43 before displayed in the console.

IIRC, it *is* parsing `number` correctly, however, it is being displayed as a rounded value when printed. Insert a breakpoint and confirm this. I may be wrong, though, hence the comment. — Tyler Crompton, Jan 15 '13 at 17:47
Please see if previous answers help - http://stackoverflow.com/questions/tagged/c%23+double?sort=faq&pagesize=50 ... — Alexei Levenkov, Jan 15 '13 at 17:48
@DStanley i need to pass the value as double to a library to make other calculations — Giorgos Manoltzas, Jan 15 '13 at 17:56
And is passing in a value that is 0.000000000007 different going to make a material difference in those calculations? If not just use a double. — D Stanley, Jan 15 '13 at 17:57
@GiorgosManoltzas - That library isn't going to be able to be any more precise with a `double` than your code will be. — Bobson, Jan 15 '13 at 18:00
That's what i though. But it is the library that exports this number. It is called Gurobi if you know it. — Giorgos Manoltzas, Jan 15 '13 at 18:06

score 11 · Accepted Answer · answered Jan 15 '13 at 17:51

11

93389.429999999993 cannot be represented exactly as a 64-bit floating point number. A double can only hold 15 or 16 digits, while you have 17 digits. If you need that level of precision use a decimal instead.

(I know you say you need it as a double, but if you could explain why, there may be alternate solutions)

answered Jan 15 '13 at 17:51

D Stanley

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thanks for helping - i need to pass the exact value a double to a library to make calculations based on the value – Giorgos Manoltzas Jan 15 '13 at 17:58
@Chux - You win; Congrats! – Security Hound Jul 09 '15 at 20:52

score 6 · Answer 2 · answered Jan 15 '13 at 17:47

This is expected behavior.

A double can't represent every number exactly. This has nothing to do with the string conversion.

You can check it yourself:

Console.WriteLine(93389.429999999993);

This will print 93389.43.

The following also shows this:

Console.WriteLine(93389.429999999993 == 93389.43);

This prints True.

score 4 · Answer 3 · answered Jan 15 '13 at 17:48

Keep in mind that there are two conversions going on here. First you're converting the string to a double, and then you're converting that double back into a string to display it.

You also need to consider that a double doesn't have infinite precision; depending on the string, some data may be lost due to the fact that a double doesn't have the capacity to store it.

When converting to a double it's not going to "round" any more than it has to. It will create the double that is closest to the number provided, given the capabilities of a double. When converting that double to a string it's much more likely that some information isn't kept.

score 3 · Answer 4 · edited May 23 '17 at 11:45

3

See the following (in particular the first part of Michael Borgwardt's answer):

decimal vs double! - Which one should I use and when?

A double will not always keep the precision depending on the number you are trying to convert

If you need to be precise you will need to use decimal

edited May 23 '17 at 11:45

Community

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answered Jan 15 '13 at 17:50

NinjaNye

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score 0 · Answer 5 · answered Jan 15 '13 at 17:50

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This is a limit on the precision that a double can store. You can see this yourself by trying to convert 3389.429999999993 instead.

answered Jan 15 '13 at 17:50

Bobson

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score 0 · Answer 6 · answered Jan 15 '13 at 18:04

The double type has a finite precision of 64 bits, so a rounding error occurs when the real number is stored in the numberAsDouble variable.

A solution that would work for your example is to use the decimal type instead, which has 128 bit precision. However, the same problem arises with a smaller difference.

For arbitrary large numbers, the System.Numerics.BigInteger object from the .NET Framework 4.0 supports arbitrary precision for integers. However you will need a 3rd party library to use arbitrary large real numbers.

I'm not sure what you mean by number, but in more details (using google) the decimal precision sign=1 bit, mantissa=96 bit, exponent=29 bit? I haven't read anything clear on the exponent, and the total # of bits used is 128 bit. — J-Mik, Jan 15 '13 at 19:59

Esteban Elverdin · Answer 7 · 2013-01-15T18:19:54.253

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You could truncate the decimal places to the amount of digits you need, not exceeding double precision.

For instance, this will truncate to 5 decimal places, getting 93389.42999. Just replace 100000 for the needed value

string number = "93389.429999999993";
decimal numberAsDecimal = Convert.ToDecimal(number);
var numberAsDouble = ((double)((long)(numberAsDecimal * 100000.0m))) / 100000.0;

edited Jan 15 '13 at 18:19

answered Jan 15 '13 at 18:07

Esteban Elverdin

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C# Convert.ToDouble() loses decimal points when converting string to double

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