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Enumerable.Min Method (IEnumerable<Double>)

Returns the minimum value in a sequence of Double values.

Namespace:  System.Linq
Assembly:  System.Core (in System.Core.dll)

public static double Min(
	this IEnumerable<double> source


Type: System.Collections.Generic.IEnumerable<Double>

A sequence of Double values to determine the minimum value of.

Return Value

Type: System.Double
The minimum value in the sequence.

Usage Note

In Visual Basic and C#, you can call this method as an instance method on any object of type IEnumerable<Double>. When you use instance method syntax to call this method, omit the first parameter. For more information, see Extension Methods (Visual Basic) or Extension Methods (C# Programming Guide).


source is null.


source contains no elements.

The Min(IEnumerable<Double>) method uses the Double implementation of IComparable<T> to compare values.

In Visual Basic query expression syntax, an Aggregate Into Min() clause translates to an invocation of Min.

The following code example demonstrates how to use Min(IEnumerable<Double>) to determine the minimum value in a sequence.

double[] doubles = { 1.5E+104, 9E+103, -2E+103 };

double min = doubles.Min();

Console.WriteLine("The smallest number is {0}.", min);

 This code produces the following output:

 The smallest number is -2E+103.

Windows 7, Windows Vista, Windows XP SP2, Windows Server 2008 R2, Windows Server 2008, Windows Server 2003, Windows CE, Windows Mobile for Smartphone, Windows Mobile for Pocket PC, Xbox 360, Zune

The .NET Framework and .NET Compact Framework do not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.

.NET Framework

Supported in: 3.5

.NET Compact Framework

Supported in: 3.5

XNA Framework

Supported in: 3.0