Why does Math.Round(2.5) return 2 instead of 3?

Question

In C#, the result of Math.Round(2.5) is 2.

It is supposed to be 3, isn\'t it? Why is it 2 instead in C#?

Answer 1

From MSDN:

By default, Math.Round uses MidpointRounding.ToEven. Most people are not familiar with "rounding to even" as the alternative, "rounding away from zero" is more commonly taught in school. .NET defaults to "Rounding to even" as it is statistically superior because it doesn't share the tendency of "rounding away from zero" to round up slightly more often than it rounds down (assuming the numbers being rounded tend to be positive.)

http://msdn.microsoft.com/en-us/library/system.math.round.aspx

Answer 2

This post has the answer you are looking for:

http://weblogs.asp.net/sfurman/archive/2003/03/07/3537.aspx

Basically this is what it says:

Return Value

The number nearest value with precision equal to digits. If value is halfway between two numbers, one of which is even and the other odd, then the even number is returned. If the precision of value is less than digits, then value is returned unchanged.

The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding to nearest, or banker's rounding. If digits is zero, this kind of rounding is sometimes called rounding toward zero.

Answer 3

Here's the way i had to work it around :

Public Function Round(number As Double, dec As Integer) As Double
    Dim decimalPowerOfTen = Math.Pow(10, dec)
    If CInt(number * decimalPowerOfTen) = Math.Round(number * decimalPowerOfTen, 2) Then
        Return Math.Round(number, 2, MidpointRounding.AwayFromZero)
    Else
        Return CInt(number * decimalPowerOfTen + 0.5) / 100
    End If
End Function

Trying with 1.905 with 2 decimals will give 1.91 as expected but Math.Round(1.905,2,MidpointRounding.AwayFromZero) gives 1.90! Math.Round method is absolutely inconsistent and unusable for most of the basics problems programmers may encounter. I have to check if (int) 1.905 * decimalPowerOfTen = Math.Round(number * decimalPowerOfTen, 2) cause i don not want to round up what should be round down.

Answer 4

That's called rounding to even (or banker's rounding), which is a valid rounding strategy for minimizing accrued errors in sums (MidpointRounding.ToEven). The theory is that, if you always round a 0.5 number in the same direction, the errors will accrue faster (round-to-even is supposed to minimize that) ^(a).

Follow these links for the MSDN descriptions of:

• Math.Floor, which rounds down towards negative infinity.
• Math.Ceiling, which rounds up towards positive infinity.
• Math.Truncate, which rounds up or down towards zero.
• Math.Round, which rounds to the nearest integer or specified number of decimal places. You can specify the behavior if it's exactly equidistant between two possibilities, such as rounding so that the final digit is even ("Round(2.5,MidpointRounding.ToEven)" becoming 2) or so that it's further away from zero ("Round(2.5,MidpointRounding.AwayFromZero)" becoming 3).

The following diagram and table may help:

-3        -2        -1         0         1         2         3
 +--|------+---------+----|----+--|------+----|----+-------|-+
    a                     b       c           d            e

                       a=-2.7  b=-0.5  c=0.3  d=1.5  e=2.8
                       ======  ======  =====  =====  =====
Floor                    -3      -1      0      1      2
Ceiling                  -2       0      1      2      3
Truncate                 -2       0      0      1      2
Round(ToEven)            -3       0      0      2      3
Round(AwayFromZero)      -3      -1      0      2      3

Note that Round is a lot more powerful than it seems, simply because it can round to a specific number of decimal places. All the others round to zero decimals always. For example:

n = 3.145;
a = System.Math.Round (n, 2, MidpointRounding.ToEven);       // 3.14
b = System.Math.Round (n, 2, MidpointRounding.AwayFromZero); // 3.15

With the other functions, you have to use multiply/divide trickery to achieve the same effect:

c = System.Math.Truncate (n * 100) / 100;                    // 3.14
d = System.Math.Ceiling (n * 100) / 100;                     // 3.15

---

^(a) Of course, that theory depends on the fact that your data has an fairly even spread of values across the even halves (0.5, 2.5, 4.5, ...) and odd halves (1.5, 3.5, ...).

If all the "half-values" are evens (for example), the errors will accumulate just as fast as if you always rounded up.

Answer 5

Firstly, this wouldn't be a C# bug anyway - it would be a .NET bug. C# is the language - it doesn't decide how Math.Round is implemented.

And secondly, no - if you read the docs, you'll see that the default rounding is "round to even" (banker's rounding):

Return Value
Type: System.Double
The integer nearest a. If the fractional component of a is halfway between two integers, one of which is even and the other odd, then the even number is returned. Note that this method returns a Double instead of an integral type.

Remarks
The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding to nearest, or banker's rounding. It minimizes rounding errors that result from consistently rounding a midpoint value in a single direction.

You can specify how Math.Round should round mid-points using an overload which takes a MidpointRounding value. There's one overload with a MidpointRounding corresponding to each of the overloads which doesn't have one:

• Round(Decimal) / Round(Decimal, MidpointRounding)
• Round(Double) / Round(Double, MidpointRounding)
• Round(Decimal, Int32) / Round(Decimal, Int32, MidpointRounding)
• Round(Double, Int32) / Round(Double, Int32, MidpointRounding)

Whether this default was well chosen or not is a different matter. (MidpointRounding was only introduced in .NET 2.0. Before then I'm not sure there was any easy way of implementing the desired behaviour without doing it yourself.) In particular, history has shown that it's not the expected behaviour - and in most cases that's a cardinal sin in API design. I can see why Banker's Rounding is useful... but it's still a surprise to many.

You may be interested to take a look at the nearest Java equivalent enum (RoundingMode) which offers even more options. (It doesn't just deal with midpoints.)

Answer 6

From MSDN, Math.Round(double a) returns:

The integer nearest a. If the fractional component of a is halfway between two integers, one of which is even and the other odd, then the even number is returned.

... and so 2.5, being halfway between 2 and 3, is rounded down to the even number (2). this is called Banker's Rounding (or round-to-even), and is a commonly-used rounding standard.

Same MSDN article:

The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding is sometimes called rounding to nearest, or banker's rounding. It minimizes rounding errors that result from consistently rounding a midpoint value in a single direction.

You can specify a different rounding behavior by calling the overloads of Math.Round that take a MidpointRounding mode.