System.Double.Equals method

Th Double.Equals(Double) method implements the System.IEquatable<T> interface, and performs slightly better than Double.Equals(Object) because it doesn't have to convert the obj parameter to an object.

Widening conversions

Depending on your programming language, it might be possible to code a Equals method where the parameter type has fewer bits (is narrower) than the instance type. This is possible because some programming languages perform an implicit widening conversion that represents the parameter as a type with as many bits as the instance.

For example, suppose the instance type is Double and the parameter type is Int32. The Microsoft C# compiler generates instructions to represent the value of the parameter as a Double object, then generates a Double.Equals(Double) method that compares the values of the instance and the widened representation of the parameter.

Consult your programming language's documentation to determine if its compiler performs implicit widening conversions of numeric types. For more information, see the Type Conversion Tables topic.

Precision in comparisons

The Equals method should be used with caution, because two apparently equivalent values can be unequal due to the differing precision of the two values. The following example reports that the Double value .333333 and the Double value returned by dividing 1 by 3 are unequal.

// Initialize two doubles with apparently identical values
double double1 = .33333;
double double2 = (double) 1/3;
// Compare them for equality
Console.WriteLine(double1.Equals(double2));    // displays false

// Initialize two doubles with apparently identical values
let double1 = 0.33333
let double2 = double (1 / 3)
// Compare them for equality
printfn $"{double1.Equals double2}"    // displays false

' Initialize two doubles with apparently identical values
Dim double1 As Double = .33333
Dim double2 As Double = 1/3
' Compare them for equality
Console.WriteLine(double1.Equals(double2))    ' displays False

Rather than comparing for equality, one technique involves defining an acceptable relative margin of difference between two values (such as .001% of one of the values). If the absolute value of the difference between the two values is less than or equal to that margin, the difference is likely to be due to differences in precision and, therefore, the values are likely to be equal. The following example uses this technique to compare .33333 and 1/3, the two Double values that the previous code example found to be unequal. In this case, the values are equal.

// Initialize two doubles with apparently identical values
double double1 = .333333;
double double2 = (double) 1/3;
// Define the tolerance for variation in their values
double difference = Math.Abs(double1 * .00001);

// Compare the values
// The output to the console indicates that the two values are equal
if (Math.Abs(double1 - double2) <= difference)
   Console.WriteLine("double1 and double2 are equal.");
else
   Console.WriteLine("double1 and double2 are unequal.");

// Initialize two doubles with apparently identical values
let double1 = 0.333333
let double2 = double (1 / 3)
// Define the tolerance for variation in their values
let difference = abs (double1 * 0.00001)

// Compare the values
// The output to the console indicates that the two values are equal
if abs (double1 - double2) <= difference then
    printfn "double1 and double2 are equal."
else
    printfn "double1 and double2 are unequal."

' Initialize two doubles with apparently identical values
Dim double1 As Double = .33333
Dim double2 As Double = 1/3
' Define the tolerance for variation in their values
Dim difference As Double = Math.Abs(double1 * .00001)

' Compare the values
' The output to the console indicates that the two values are equal
If Math.Abs(double1 - double2) <= difference Then
   Console.WriteLine("double1 and double2 are equal.")
Else
   Console.WriteLine("double1 and double2 are unequal.")
End If

Note

Because Epsilon defines the minimum expression of a positive value whose range is near zero, the margin of difference between two similar values must be greater than Epsilon. Typically, it is many times greater than Epsilon. Because of this, we recommend that you don't use Epsilon when comparing Double values for equality.

A second technique involves comparing the difference between two floating-point numbers with some absolute value. If the difference is less than or equal to that absolute value, the numbers are equal. If it's greater, the numbers are not equal. One alternative is to arbitrarily select an absolute value. That's problematic, however, because an acceptable margin of difference depends on the magnitude of the Double values. A second alternative takes advantage of a design feature of the floating-point format: The difference between the integer representation of two floating-point values indicates the number of possible floating-point values that separates them. For example, the difference between 0.0 and Epsilon is 1, because Epsilon is the smallest representable value when working with a Double whose value is zero. The following example uses this technique to compare .33333 and 1/3, which are the two Double values that the previous code example with the Equals(Double) method found to be unequal. The example uses the BitConverter.DoubleToInt64Bits method to convert a double-precision floating-point value to its integer representation. The example declares the values as equal if there are no possible floating-point values between their integer representations.

public static void Main()
{
    // Initialize the values.
    double value1 = .1 * 10;
    double value2 = 0;
    for (int ctr = 0; ctr < 10; ctr++)
        value2 += .1;

    Console.WriteLine($"{value1:R} = {value2:R}: " +
        $"{HasMinimalDifference(value1, value2, 1)}");
}

public static bool HasMinimalDifference(
    double value1,
    double value2,
    int allowableDifference
    )
{
    // Convert the double values to long values.
    long lValue1 = BitConverter.DoubleToInt64Bits(value1);
    long lValue2 = BitConverter.DoubleToInt64Bits(value2);

    // If the signs are different, return false except for +0 and -0.
    if ((lValue1 >> 63) != (lValue2 >> 63))
    {
        if (value1 == value2)
            return true;

        return false;
    }

    // Calculate the number of possible
    // floating-point values in the difference.
    long diff = Math.Abs(lValue1 - lValue2);

    if (diff <= allowableDifference)
        return true;

    return false;
}
// The example displays the following output:
//
//        1 = 0.99999999999999989: True

open System

let hasMinimalDifference (value1: double) (value2: double) (units: int) =
    let lValue1 = BitConverter.DoubleToInt64Bits value1
    let lValue2 = BitConverter.DoubleToInt64Bits value2

    // If the signs are different, return false except for +0 and -0.
    if (lValue1 >>> 63) <> (lValue2 >>> 63) then
        value1 = value2
    else
        let diff = abs (lValue1 - lValue2)

        diff <= int64 units

let value1 = 0.1 * 10.
let mutable value2 = 0.
for _ = 0 to 9 do
    value2 <- value2 + 0.1

printfn $"{value1:R} = {value2:R}: {hasMinimalDifference value1 value2 1}"
                

// The example displays the following output:
//        1 = 0.99999999999999989: True

Module Example
   Public Sub Main()
      Dim value1 As Double = .1 * 10
      Dim value2 As Double = 0
      For ctr As Integer =  0 To 9
         value2 += .1
      Next
               
      Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2,
                        HasMinimalDifference(value1, value2, 1))
   End Sub

   Public Function HasMinimalDifference(value1 As Double, value2 As Double, units As Integer) As Boolean
      Dim lValue1 As long =  BitConverter.DoubleToInt64Bits(value1)
      Dim lValue2 As long =  BitConverter.DoubleToInt64Bits(value2)
      
      ' If the signs are different, Return False except for +0 and -0.
      If ((lValue1 >> 63) <> (lValue2 >> 63)) Then
         If value1 = value2 Then
            Return True
         End If           
         Return False
      End If

      Dim diff As Long =  Math.Abs(lValue1 - lValue2)

      If diff <= units Then
         Return True
      End If

      Return False
   End Function
End Module
' The example displays the following output:
'       1 = 0.99999999999999989: True

Note

For some values, you might consider them equal even when there is a possible floating-point value between the integer representations. For example, consider the double values 0.39 and 1.69 - 1.3 (which is calculated as 0.3899999999999999). On a little-endian computer, the integer representations of these values are 4600697235336603894 and 4600697235336603892, respectively. The difference between the integer values is 2, meaning there is a possible floating-point value between 0.39 and 1.69 - 1.3.

Version differences

The precision of floating-point numbers beyond the documented precision is specific to the implementation and version of .NET. Consequently, a comparison of two particular numbers might change between versions of .NET because the precision of the numbers' internal representation might change.

NaN

If two Double.NaN values are tested for equality by calling the Equals method, the method returns true. However, if two Double.NaN values are tested for equality by using the equality operator, the operator returns false. When you want to determine whether the value of a Double is not a number (NaN), an alternative is to call the IsNaN method.