Rounding to specfic digits fails with this double-precision value

Question

I'm attempting to truncate a series of double-precision values in C#. The following value fails no matter what rounding method I use. What is wrong with this value that causes both of these methods to fail? Why does even Math.Round fail to correctly truncate the number? What method can be used instead to correctly truncate such values?

The value :

double value = 0.61740451388888251;

Method 1:

return Math.Round(value, digits);

Method 2:

double multiplier = Math.Pow(10, decimals)
return Math.Round(value * multiplier) / multiplier;

Fails even in VS watch window!

round

Floating-point types can't represent all possible values, meaning this is probably the closest approximation of 0.62 possible. — Jordi Vermeulen, Dec 05 '14 at 19:45
@JordiVermeulen - `double` can surely represent 0.62 correctly. — Robin Rodricks, Dec 05 '14 at 19:47

PiotrWolkowski · Accepted Answer · 2014-12-05T20:42:40.633

Double is a floating binary point type. They are represented in binary system (like 11010.00110). When double is presented in decimal system it is only an approximation as not all binary numbers have exact representation in decimal system. Try for example this operation:

double d = 3.65d + 0.05d;

It will not result in 3.7 but in 3.6999999999999997. It is because the variable contains a closest available double.

The same happens in your case. Your variable contains closest available double.

For precise operations double/float is not the most fortunate choice. Use double/float when you need fast performance or you want to operate on larger range of numbers, but where high precision is not required. For instance, it is perfect type for calculations in physics. For precise decimal operations use, well, decimal.

Here is an article about float/decimal: http://csharpindepth.com/Articles/General/FloatingPoint.aspx

eigenchris · Answer 2 · 2014-12-05T20:06:43.137

3

According to this online tool which gives the binary representation of doubles, the two closest double values to 0.62 are:

6.19999999999999995559107901499E-1 or 0x3FE3D70A3D70A3D7 link

6.20000000000000106581410364015E-1 or 0x3FE3D70A3D70A3D8 link

I'm not sure why neither of these agree with your value exactly, but like the others said, it is likely a floating point representation issue.

edited Dec 05 '14 at 20:06

answered Dec 05 '14 at 19:58

eigenchris

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score 3 · Answer 3 · edited May 23 '17 at 12:28

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If you need a more exact representation of the number you might have to use the decimal type, which has more precision but smaller range (it's usually used financial calculations).

More info on when to use each here: https://stackoverflow.com/a/618596/1373170

edited May 23 '17 at 12:28

Community

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answered Dec 05 '14 at 20:03

Pablo Romeo

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score -1 · Answer 4 · answered Dec 05 '14 at 19:48

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I think you are running up against the binary limit of a double-precision float (64 bits). From http://en.wikipedia.org/wiki/Double-precision_floating-point_format, a double only gives between 15-17 significant digits.

answered Dec 05 '14 at 19:48

ryanyuyu

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Why does `Math.Round` fail to correctly truncate the number? What method can be used instead? – Robin Rodricks Dec 05 '14 at 19:58
Are you just trying to round to 2 decimal places? – ryanyuyu Dec 05 '14 at 20:02
Since you are using C#, you can use the Decimal class. http://msdn.microsoft.com/en-us/library/System.Decimal(v=vs.110).aspx – ryanyuyu Dec 05 '14 at 20:07

Rounding to specfic digits fails with this double-precision value

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