Enumerable.Last 方法

定义

命名空间:
System.Linq
程序集:
System.Linq.dll, System.Core.dll, netstandard.dll

返回序列的最后一个元素。Returns the last element of a sequence.

重载

Last<TSource>(IEnumerable<TSource>)

返回序列的最后一个元素。Returns the last element of a sequence.
Last<TSource>(IEnumerable<TSource>, Func<TSource,Boolean>)

返回序列中满足指定条件的最后一个元素。Returns the last element of a sequence that satisfies a specified condition.

Last<TSource>(IEnumerable<TSource>)

返回序列的最后一个元素。Returns the last element of a sequence.

public:
generic <typename TSource>
[System::Runtime::CompilerServices::Extension]
 static TSource Last(System::Collections::Generic::IEnumerable<TSource> ^ source);
public static TSource Last<TSource> (this System.Collections.Generic.IEnumerable<TSource> source);
static member Last : seq<'Source> -> 'Source
<Extension()>
Public Function Last(Of TSource) (source As IEnumerable(Of TSource)) As TSource

类型参数

TSource

source 的元素类型。The type of the elements of source.

参数

source
IEnumerable<TSource>

要返回其最后一个元素的 IEnumerable<T>。An IEnumerable<T> to return the last element of.

返回

TSource

源序列中最后位置处的值。The value at the last position in the source sequence.

例外

ArgumentNullException

source 为 null。source is null.

InvalidOperationException

源序列为空。The source sequence is empty.

示例

下面的代码示例演示如何使用 Last<TSource>(IEnumerable<TSource>) 返回数组的最后一个元素。The following code example demonstrates how to use Last<TSource>(IEnumerable<TSource>) to return the last element of an array.

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

int last = numbers.Last();

Console.WriteLine(last);

/*
 This code produces the following output:

 19
*/

' 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 item in the array.
Dim last As Integer = numbers.Last()

' Display the result.
Console.WriteLine(last)

' This code produces the following output:
'
' 19

注解

如果 source 不包含任何元素，则 Last<TSource>(IEnumerable<TSource>) 方法会引发异常。The Last<TSource>(IEnumerable<TSource>) method throws an exception if source contains no elements. 若要改为在源序列为空时返回默认值，请使用 LastOrDefault 方法。To instead return a default value when the source sequence is empty, use the LastOrDefault method.

Last<TSource>(IEnumerable<TSource>, Func<TSource,Boolean>)

返回序列中满足指定条件的最后一个元素。Returns the last element of a sequence that satisfies a specified condition.

public:
generic <typename TSource>
[System::Runtime::CompilerServices::Extension]
 static TSource Last(System::Collections::Generic::IEnumerable<TSource> ^ source, Func<TSource, bool> ^ predicate);
public static TSource Last<TSource> (this System.Collections.Generic.IEnumerable<TSource> source, Func<TSource,bool> predicate);
static member Last : seq<'Source> * Func<'Source, bool> -> 'Source
<Extension()>
Public Function Last(Of TSource) (source As IEnumerable(Of TSource), predicate As Func(Of TSource, Boolean)) As TSource

类型参数

TSource

source 的元素类型。The type of the elements of source.

参数

source
IEnumerable<TSource>

要从中返回元素的 IEnumerable<T>。An IEnumerable<T> to return an element from.

predicate
Func<TSource,Boolean>

用于测试每个元素是否满足条件的函数。A function to test each element for a condition.

返回

TSource

序列中通过指定谓词函数中的测试的最后一个元素。The last element in the sequence that passes the test in the specified predicate function.

例外

ArgumentNullException

source 或 predicate 为 null。source or predicate is null.

InvalidOperationException

元素均不满足 predicate 中的条件。No element satisfies the condition in predicate.

- 或 --or- 源序列为空。The source sequence is empty.

示例

下面的代码示例演示如何使用 Last<TSource>(IEnumerable<TSource>, Func<TSource,Boolean>) 返回数组中满足条件的最后一个元素。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.

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

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

Console.WriteLine(last);

/*
 This code produces the following output:

 87
*/

' 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.
Console.WriteLine(last)

' This code produces the following output:
'
' 87

注解

如果 source中找不到匹配的元素，Last<TSource>(IEnumerable<TSource>, Func<TSource,Boolean>) 方法将引发异常。The Last<TSource>(IEnumerable<TSource>, Func<TSource,Boolean>) method throws an exception if no matching element is found in source. 如果未找到匹配的元素，则改为返回默认值时，请使用 LastOrDefault 方法。To instead return a default value when no matching element is found, use the LastOrDefault method.

适用于