# System.Double.CompareTo methods

This article provides supplementary remarks to the reference documentation for this API.

## CompareTo(Double) method

Values must be identical to be considered equal. Particularly when floating-point values depend on multiple mathematical operations, it is common for them to lose precision and for their values to be nearly identical except for their least significant digits. Because of this, the return value of the CompareTo method at times may seem surprising. For example, multiplication by a particular value followed by division by the same value should produce the original value. In the following example, however, the computed value turns out to be greater than the original value. Showing all significant digits of the two values by using the "R" standard numeric format string indicates that the computed value differs from the original value in its least significant digits. For information on handling such comparisons, see the Remarks section of the Equals(Double) method.

```
using System;
public class Example
{
public static void Main()
{
double value1 = 6.185;
double value2 = value1 * .1 / .1;
Console.WriteLine("Comparing {0} and {1}: {2}\n",
value1, value2, value1.CompareTo(value2));
Console.WriteLine("Comparing {0:R} and {1:R}: {2}",
value1, value2, value1.CompareTo(value2));
}
}
// The example displays the following output:
// Comparing 6.185 and 6.185: -1
//
// Comparing 6.185 and 6.1850000000000005: -1
```

```
let value1 = 6.185
let value2 = value1 * 0.1 / 0.1
printfn $"Comparing {value1} and {value2}: {value1.CompareTo value2}\n"
printfn $"Comparing {value1:R} and {value2:R}: {value1.CompareTo value2}"
// The example displays the following output:
// Comparing 6.185 and 6.185: -1
//
// Comparing 6.185 and 6.1850000000000005: -1
```

```
Module Example
Public Sub Main()
Dim value1 As Double = 6.185
Dim value2 As Double = value1 * .1 / .1
Console.WriteLine("Comparing {0} and {1}: {2}",
value1, value2, value1.CompareTo(value2))
Console.WriteLine()
Console.WriteLine("Comparing {0:R} and {1:R}: {2}",
value1, value2, value1.CompareTo(value2))
End Sub
End Module
' The example displays the following output:
' Comparing 6.185 and 6.185: -1
'
' Comparing 6.185 and 6.1850000000000005: -1
```

This method implements the System.IComparable<T> interface and performs slightly better than the Double.CompareTo method because it does not have to convert the `value`

parameter to an object.

Note that, although an object whose value is NaN is not considered equal to another object whose value is NaN (even itself), the IComparable<T> interface requires that `A.CompareTo(A)`

return zero.

## CompareTo(Object) method

The `value`

parameter must be `null`

or an instance of `Double`

; otherwise, an exception is thrown. Any instance of Double, regardless of its value, is considered greater than `null`

.

Values must be identical to be considered equal. Particularly when floating-point values depend on multiple mathematical operations, it is common for them to lose precision and for their values to be nearly identical except for their least significant digits. Because of this, the return value of the CompareTo method at times may seem surprising. For example, multiplication by a particular value followed by division by the same value should produce the original value. In the following example, however, the computed value turns out to be greater than the original value. Showing all significant digits of the two values by using the "R" standard numeric format string indicates that the computed value differs from the original value in its least significant digits. For information on handling such comparisons, see the Remarks section of the Equals(Double) method.

```
using System;
public class Example3
{
public static void Main()
{
double value1 = 6.185;
object value2 = value1 * .1 / .1;
Console.WriteLine("Comparing {0} and {1}: {2}\n",
value1, value2, value1.CompareTo(value2));
Console.WriteLine("Comparing {0:R} and {1:R}: {2}",
value1, value2, value1.CompareTo(value2));
}
}
// The example displays the following output:
// Comparing 6.185 and 6.185: -1
//
// Comparing 6.185 and 6.1850000000000005: -1
```

```
let value1 = 6.185
let value2 = value1 * 0.1 / 0.1 |> box
printfn $"Comparing {value1} and {value2}: {value1.CompareTo value2}\n"
printfn $"Comparing {value1:R} and {value2:R}: {value1.CompareTo value2}"
// The example displays the following output:
// Comparing 6.185 and 6.185: -1
//
// Comparing 6.185 and 6.1850000000000005: -1
```

```
Module Example2
Public Sub Main()
Dim value1 As Double = 6.185
Dim value2 As Object = value1 * .1 / .1
Console.WriteLine("Comparing {0} and {1}: {2}",
value1, value2, value1.CompareTo(value2))
Console.WriteLine()
Console.WriteLine("Comparing {0:R} and {1:R}: {2}",
value1, value2, value1.CompareTo(value2))
End Sub
End Module
' The example displays the following output:
' Comparing 6.185 and 6.185: -1
'
' Comparing 6.185 and 6.1850000000000005: -1
```

This method is implemented to support the IComparable interface. Note that, although a NaN is not considered to be equal to another NaN (even itself), the IComparable interface requires that `A.CompareTo(A)`

return zero.

## Widening conversions

Depending on your programming language, it might be possible to code a CompareTo 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.CompareTo(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 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.

## Feedback

https://aka.ms/ContentUserFeedback.

Coming soon: Throughout 2024 we will be phasing out GitHub Issues as the feedback mechanism for content and replacing it with a new feedback system. For more information see:Submit and view feedback for