Поделиться через


Enumerable.ElementAt<TSource> Method

Microsoft Silverlight will reach end of support after October 2021. Learn more.

Returns the element at a specified index in a sequence.

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

Syntax

'Declaration
<ExtensionAttribute> _
Public Shared Function ElementAt(Of TSource) ( _
    source As IEnumerable(Of TSource), _
    index As Integer _
) As TSource
public static TSource ElementAt<TSource>(
    this IEnumerable<TSource> source,
    int index
)

Type Parameters

  • TSource
    The type of the elements of source.

Parameters

  • index
    Type: System.Int32
    The zero-based index of the element to retrieve.

Return Value

Type: TSource
The element at the specified position in the source sequence.

Usage Note

In Visual Basic and C#, you can call this method as an instance method on any object of type IEnumerable<TSource>. When you use instance method syntax to call this method, omit the first parameter.

Exceptions

Exception Condition
ArgumentNullException

source is nulla null reference (Nothing in Visual Basic).

ArgumentOutOfRangeException

index is less than 0 or greater than or equal to the number of elements in source.

Remarks

If the type of source implements IList<T>, that implementation is used to obtain the element at the specified index. Otherwise, this method obtains the specified element.

This method throws an exception if index is out of range. To instead return a default value when the specified index is out of range, use the ElementAtOrDefault<TSource> method.

Examples

The following code example demonstrates how to use ElementAt<TSource> to return an element at a specific position.

      ' Create an array of strings.
      Dim names() As String = _
          {"Hartono, Tommy", "Adams, Terry", "Andersen, Henriette Thaulow", "Hedlund, Magnus", "Ito, Shu"}

      Dim random As Random = New Random(DateTime.Now.Millisecond)

      ' Get a string at a random index within the array.
      Dim name As String = names.ElementAt(random.Next(0, names.Length))

      ' Display the output.
      outputBlock.Text &= "The name chosen at random is " & name & vbCrLf

      ' This code produces the following output:
      '
      ' The name chosen at random is Ito, Shu

      string[] names =
             { "Hartono, Tommy", "Adams, Terry", "Andersen, Henriette Thaulow", 

                 "Hedlund, Magnus", "Ito, Shu" };
      Random random = new Random(DateTime.Now.Millisecond);

      string name = names.ElementAt(random.Next(0, names.Length));

      outputBlock.Text += String.Format("The name chosen at random is '{0}'.", name) + "\n";

      /*
       This code produces the following sample output:

       The name chosen at random is 'Ito, Shu'.
      */

Version Information

Silverlight

Supported in: 5, 4, 3

Silverlight for Windows Phone

Supported in: Windows Phone OS 7.1, Windows Phone OS 7.0

XNA Framework

Supported in: Xbox 360, Windows Phone OS 7.0

Platforms

For a list of the operating systems and browsers that are supported by Silverlight, see Supported Operating Systems and Browsers.