Share via


Enumerable.Last<TSource> Method (IEnumerable<TSource>, Func<TSource, Boolean>)

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

Returns the last element of a sequence that satisfies a specified condition.

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

Syntax

'Declaration
<ExtensionAttribute> _
Public Shared Function Last(Of TSource) ( _
    source As IEnumerable(Of TSource), _
    predicate As Func(Of TSource, Boolean) _
) As TSource
public static TSource Last<TSource>(
    this IEnumerable<TSource> source,
    Func<TSource, bool> predicate
)

Type Parameters

  • TSource
    The type of the elements of source.

Parameters

  • predicate
    Type: System.Func<TSource, Boolean>
    A function to test each element for a condition.

Return Value

Type: TSource
The last element in the sequence that passes the test in the specified predicate function.

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 or predicate is nulla null reference (Nothing in Visual Basic).

InvalidOperationException

No element satisfies the condition in predicate.

-or-

The source sequence is empty.

Remarks

The Last<TSource>(IEnumerable<TSource>, Func<TSource, Boolean>) method throws an exception if no matching element is found in source. To instead return a default value when no matching element is found, use the LastOrDefault method.

Examples

The following code example demonstrates how to use Last<TSource>(IEnumerable<TSource>, Func<TSource, Boolean>) to return the last element of an array that satisfies a condition.

      ' Create an array of integers.
      Dim numbers() As Integer = _
          {9, 34, 65, 92, 87, 435, 3, 54, 83, 23, 87, 67, 12, 19}

      ' Get the last element in the array whose value is
      ' greater than 80.
      Dim last As Integer = numbers.Last(Function(num) num > 80)

      ' Display the result.
      outputBlock.Text &= last & vbCrLf

      ' This code produces the following output:
      '
      ' 87

      int[] numbers = { 9, 34, 65, 92, 87, 435, 3, 54, 
                             83, 23, 87, 67, 12, 19 };

      int last = numbers.Last(num => num > 80);

      outputBlock.Text += last + "\n";

      /*
       This code produces the following output:

       87
      */

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.