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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
)

Parameters

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.

Exception Condition
ArgumentNullException

source is null.

InvalidOperationException

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.
*/

Universal Windows Platform
Available since 8
.NET Framework
Available since 3.5
Portable Class Library
Supported in: portable .NET platforms
Silverlight
Available since 2.0
Windows Phone Silverlight
Available since 7.0
Windows Phone
Available since 8.1
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