Min Method (IEnumerable(Double))

Enumerable.Min Method (IEnumerable<Double>)

[ This article is for Windows Phone 8 developers. If you’re developing for Windows 10, see the latest documentation. ]

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.

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.

ExceptionCondition
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 Enumerable.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();

      outputBlock.Text += String.Format("The smallest number is {0}.", min) + "\n";

      /*
       This code produces the following output:

       The smallest number is -2E+103.
      */



Windows Phone OS

Supported in: 8.1, 8.0, 7.1, 7.0

Windows Phone

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