Array.LastIndexOf 方法

定义

返回一维 ArrayArray 的一部分中某个值的最后一个匹配项的索引。Returns the index of the last occurrence of a value in a one-dimensional Array or in a portion of the Array.

重载

LastIndexOf(Array, Object)

在整个一维 Array 中搜索指定的对象,并返回最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the entire one-dimensional Array.

LastIndexOf(Array, Object, Int32)

搜索指定的对象,并返回一维 Array 中从第一个元素到指定索引的元素范围内最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the one-dimensional Array that extends from the first element to the specified index.

LastIndexOf(Array, Object, Int32, Int32)

搜索指定的对象并返回一维 Array 中包含指定数目元素且在指定索引处结尾的元素范围内的最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the one-dimensional Array that contains the specified number of elements and ends at the specified index.

LastIndexOf<T>(T[], T)

搜索指定的对象,并返回整个 Array 中最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the entire Array.

LastIndexOf<T>(T[], T, Int32)

搜索指定的对象并返回 Array 中从第一个元素到指定索引的元素范围内最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the Array that extends from the first element to the specified index.

LastIndexOf<T>(T[], T, Int32, Int32)

搜索指定的对象,并返回 Array 中包含指定数目元素并在指定索引处结尾的元素范围内的最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the Array that contains the specified number of elements and ends at the specified index.

LastIndexOf(Array, Object)

在整个一维 Array 中搜索指定的对象,并返回最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the entire one-dimensional Array.

public:
 static int LastIndexOf(Array ^ array, System::Object ^ value);
public static int LastIndexOf (Array array, object value);
static member LastIndexOf : Array * obj -> int

参数

array
Array

要搜索的一维 ArrayThe one-dimensional Array to search.

value
Object

要在 array 中查找的对象。The object to locate in array.

返回

如果在整个 value 中找到 array 的最后一个匹配项的索引,则为该索引;否则为该数组的下限减 1。The index of the last occurrence of value within the entire array, if found; otherwise, the lower bound of the array minus 1.

异常

arraynullarray is null.

array 是多维的。array is multidimensional.

示例

下面的代码示例演示如何确定指定元素在数组中的最后一个匹配项的索引。The following code example shows how to determine the index of the last occurrence of a specified element in an array.

using namespace System;
void PrintIndexAndValues( Array^ myArray );

void main()
{
   // Creates and initializes a new Array instance with three elements of the same value.
   Array^ myArray = Array::CreateInstance( String::typeid, 12 );
   myArray->SetValue( "the", 0 );
   myArray->SetValue( "quick", 1 );
   myArray->SetValue( "brown", 2 );
   myArray->SetValue( "fox", 3 );
   myArray->SetValue( "jumps", 4 );
   myArray->SetValue( "over", 5 );
   myArray->SetValue( "the", 6 );
   myArray->SetValue( "lazy", 7 );
   myArray->SetValue( "dog", 8 );
   myArray->SetValue( "in", 9 );
   myArray->SetValue( "the", 10 );
   myArray->SetValue( "barn", 11 );

   // Displays the values of the Array.
   Console::WriteLine(  "The Array instance contains the following values:" );
   PrintIndexAndValues( myArray );

   // Searches for the last occurrence of the duplicated value.
   String^ myString =  "the";
   int myIndex = Array::LastIndexOf( myArray, myString );
   Console::WriteLine(  "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in the first section of the Array.
   myIndex = Array::LastIndexOf( myArray, myString, 8 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in a section of the Array.  
   // Note that the start index is greater than the end index because the search is done backward.
   myIndex = Array::LastIndexOf( myArray, myString, 10, 6 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );
}

void PrintIndexAndValues( Array^ myArray )
{
   for ( int i = myArray->GetLowerBound( 0 ); i <= myArray->GetUpperBound( 0 ); i++ )
      Console::WriteLine(  "\t[{0}]:\t{1}", i, myArray->GetValue( i ) );
}

/* 
 This code produces the following output.
 
 The Array instance contains the following values:
     [0]:    the
     [1]:    quick
     [2]:    brown
     [3]:    fox
     [4]:    jumps
     [5]:    over
     [6]:    the
     [7]:    lazy
     [8]:    dog
     [9]:    in
     [10]:    the
     [11]:    barn
 The last occurrence of "the" is at index 10.
 The last occurrence of "the" between the start and index 8 is at index 6.
 The last occurrence of "the" between index 5 and index 10 is at index 10.
 */
// Creates and initializes a new Array with three elements of the same value.
Array myArray=Array.CreateInstance( typeof(String), 12 );
myArray.SetValue( "the", 0 );
myArray.SetValue( "quick", 1 );
myArray.SetValue( "brown", 2 );
myArray.SetValue( "fox", 3 );
myArray.SetValue( "jumps", 4 );
myArray.SetValue( "over", 5 );
myArray.SetValue( "the", 6 );
myArray.SetValue( "lazy", 7 );
myArray.SetValue( "dog", 8 );
myArray.SetValue( "in", 9 );
myArray.SetValue( "the", 10 );
myArray.SetValue( "barn", 11 );

// Displays the values of the Array.
Console.WriteLine( "The Array contains the following values:" );
PrintIndexAndValues( myArray );

// Searches for the last occurrence of the duplicated value.
String myString = "the";
int myIndex = Array.LastIndexOf( myArray, myString );
Console.WriteLine( "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in the first section of the Array.
myIndex = Array.LastIndexOf( myArray, myString, 8 );
Console.WriteLine( "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in a section of the Array.
// Note that the start index is greater than the end index because the search is done backward.
myIndex = Array.LastIndexOf( myArray, myString, 10, 6 );
Console.WriteLine( "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );

void PrintIndexAndValues( Array anArray )  {
   for ( int i = anArray.GetLowerBound(0); i <= anArray.GetUpperBound(0); i++ )
      Console.WriteLine( "\t[{0}]:\t{1}", i, anArray.GetValue( i ) );
}

/* 
This code produces the following output.

The Array contains the following values:
   [0]:    the
   [1]:    quick
   [2]:    brown
   [3]:    fox
   [4]:    jumps
   [5]:    over
   [6]:    the
   [7]:    lazy
   [8]:    dog
   [9]:    in
   [10]:    the
   [11]:    barn
The last occurrence of "the" is at index 10.
The last occurrence of "the" between the start and index 8 is at index 6.
The last occurrence of "the" between index 5 and index 10 is at index 10.
*/
Public Class SamplesArray    
    
    Public Shared Sub Main()
        
        ' Creates and initializes a new Array with three elements of
        ' the same value.
        Dim myArray As Array = Array.CreateInstance(GetType(String), 12)
        myArray.SetValue("the", 0)
        myArray.SetValue("quick", 1)
        myArray.SetValue("brown", 2)
        myArray.SetValue("fox", 3)
        myArray.SetValue("jumps", 4)
        myArray.SetValue("over", 5)
        myArray.SetValue("the", 6)
        myArray.SetValue("lazy", 7)
        myArray.SetValue("dog", 8)
        myArray.SetValue("in", 9)
        myArray.SetValue("the", 10)
        myArray.SetValue("barn", 11)
        
        ' Displays the values of the Array.
        Console.WriteLine("The Array contains the following values:")
        PrintIndexAndValues(myArray)
        
        ' Searches for the last occurrence of the duplicated value.
        Dim myString As String = "the"
        Dim myIndex As Integer = Array.LastIndexOf(myArray, myString)
        Console.WriteLine("The last occurrence of ""{0}"" is at index {1}.", _
           myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in the first
        ' section of the Array.
        myIndex = Array.LastIndexOf(myArray, myString, 8)
        Console.WriteLine("The last occurrence of ""{0}"" between the start " _
           + "and index 8 is at index {1}.", myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in a section
        ' of the Array.  Note that the start index is greater than the end
        ' index because the search is done backward.
        myIndex = Array.LastIndexOf(myArray, myString, 10, 6)
        Console.WriteLine("The last occurrence of ""{0}"" between index 5 " _
           + "and index 10 is at index {1}.", myString, myIndex)
    End Sub
    
    
    Public Shared Sub PrintIndexAndValues(myArray As Array)
        Dim i As Integer
        For i = myArray.GetLowerBound(0) To myArray.GetUpperBound(0)
            Console.WriteLine(ControlChars.Tab + "[{0}]:" + ControlChars.Tab _
               + "{1}", i, myArray.GetValue(i))
        Next i
    End Sub
End Class

' This code produces the following output.
' 
' The Array contains the following values:
'     [0]:    the
'     [1]:    quick
'     [2]:    brown
'     [3]:    fox
'     [4]:    jumps
'     [5]:    over
'     [6]:    the
'     [7]:    lazy
'     [8]:    dog
'     [9]:    in
'     [10]:    the
'     [11]:    barn
' The last occurrence of "the" is at index 10.
' The last occurrence of "the" between the start and index 8 is at index 6.
' The last occurrence of "the" between index 5 and index 10 is at index 10. 

注解

从最后一个元素开始向后搜索一维 Array,并在第一个元素处结束。The one-dimensional Array is searched backward starting at the last element and ending at the first element.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的 Equals 实现。If the element type is a nonintrinsic (user-defined) type, the Equals implementation of that type is used.

由于大多数数组的下限均为零,因此当找不到 value 时,此方法通常会返回-1。Since most arrays will have a lower bound of zero, this method would generally return -1 when value is not found. 在极少数情况下,数组的下限等于 Int32.MinValue 并且找不到 value,则此方法返回 Int32.MaxValue,这 System.Int32.MinValue - 1In the rare case that the lower bound of the array is equal to Int32.MinValue and value is not found, this method returns Int32.MaxValue, which is System.Int32.MinValue - 1.

此方法是一个 O (n)操作,其中 narrayLengthThis method is an O(n) operation, where n is the Length of array.

在 .NET Framework 版本2.0 中,此方法使用 ArrayEqualsCompareTo 方法来确定 Object 参数指定的 value 是否存在。In the .NET Framework version 2.0, this method uses the Equals and CompareTo methods of the Array to determine whether the Object specified by the value parameter exists. 在 .NET Framework 的早期版本中,此决定是使用 valueObject 本身的 EqualsCompareTo 方法进行的。In the earlier versions of the .NET Framework, this determination was made by using the Equals and CompareTo methods of the valueObject itself.

集合中对象的 item 参数的 CompareTo 方法。CompareTo methods of the item parameter on the objects in the collection.

另请参阅

LastIndexOf(Array, Object, Int32)

搜索指定的对象,并返回一维 Array 中从第一个元素到指定索引的元素范围内最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the one-dimensional Array that extends from the first element to the specified index.

public:
 static int LastIndexOf(Array ^ array, System::Object ^ value, int startIndex);
public static int LastIndexOf (Array array, object value, int startIndex);
static member LastIndexOf : Array * obj * int -> int

参数

array
Array

要搜索的一维 ArrayThe one-dimensional Array to search.

value
Object

要在 array 中查找的对象。The object to locate in array.

startIndex
Int32

向后搜索的起始索引。The starting index of the backward search.

返回

如果在 value 中从第一个元素到 array 的元素范围内找到了 startIndex 的最后一个匹配项的索引,则为该索引;否则为该数组的下限减 1。The index of the last occurrence of value within the range of elements in array that extends from the first element to startIndex, if found; otherwise, the lower bound of the array minus 1.

异常

arraynullarray is null.

startIndex 超出了 array 的有效索引范围。startIndex is outside the range of valid indexes for array.

array 是多维的。array is multidimensional.

示例

下面的代码示例演示如何确定指定元素在数组中的最后一个匹配项的索引。The following code example shows how to determine the index of the last occurrence of a specified element in an array.

using namespace System;
void PrintIndexAndValues( Array^ myArray );

void main()
{
   // Creates and initializes a new Array instance with three elements of the same value.
   Array^ myArray = Array::CreateInstance( String::typeid, 12 );
   myArray->SetValue( "the", 0 );
   myArray->SetValue( "quick", 1 );
   myArray->SetValue( "brown", 2 );
   myArray->SetValue( "fox", 3 );
   myArray->SetValue( "jumps", 4 );
   myArray->SetValue( "over", 5 );
   myArray->SetValue( "the", 6 );
   myArray->SetValue( "lazy", 7 );
   myArray->SetValue( "dog", 8 );
   myArray->SetValue( "in", 9 );
   myArray->SetValue( "the", 10 );
   myArray->SetValue( "barn", 11 );

   // Displays the values of the Array.
   Console::WriteLine(  "The Array instance contains the following values:" );
   PrintIndexAndValues( myArray );

   // Searches for the last occurrence of the duplicated value.
   String^ myString =  "the";
   int myIndex = Array::LastIndexOf( myArray, myString );
   Console::WriteLine(  "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in the first section of the Array.
   myIndex = Array::LastIndexOf( myArray, myString, 8 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in a section of the Array.  
   // Note that the start index is greater than the end index because the search is done backward.
   myIndex = Array::LastIndexOf( myArray, myString, 10, 6 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );
}

void PrintIndexAndValues( Array^ myArray )
{
   for ( int i = myArray->GetLowerBound( 0 ); i <= myArray->GetUpperBound( 0 ); i++ )
      Console::WriteLine(  "\t[{0}]:\t{1}", i, myArray->GetValue( i ) );
}

/* 
 This code produces the following output.
 
 The Array instance contains the following values:
     [0]:    the
     [1]:    quick
     [2]:    brown
     [3]:    fox
     [4]:    jumps
     [5]:    over
     [6]:    the
     [7]:    lazy
     [8]:    dog
     [9]:    in
     [10]:    the
     [11]:    barn
 The last occurrence of "the" is at index 10.
 The last occurrence of "the" between the start and index 8 is at index 6.
 The last occurrence of "the" between index 5 and index 10 is at index 10.
 */
// Creates and initializes a new Array with three elements of the same value.
Array myArray=Array.CreateInstance( typeof(String), 12 );
myArray.SetValue( "the", 0 );
myArray.SetValue( "quick", 1 );
myArray.SetValue( "brown", 2 );
myArray.SetValue( "fox", 3 );
myArray.SetValue( "jumps", 4 );
myArray.SetValue( "over", 5 );
myArray.SetValue( "the", 6 );
myArray.SetValue( "lazy", 7 );
myArray.SetValue( "dog", 8 );
myArray.SetValue( "in", 9 );
myArray.SetValue( "the", 10 );
myArray.SetValue( "barn", 11 );

// Displays the values of the Array.
Console.WriteLine( "The Array contains the following values:" );
PrintIndexAndValues( myArray );

// Searches for the last occurrence of the duplicated value.
String myString = "the";
int myIndex = Array.LastIndexOf( myArray, myString );
Console.WriteLine( "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in the first section of the Array.
myIndex = Array.LastIndexOf( myArray, myString, 8 );
Console.WriteLine( "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in a section of the Array.
// Note that the start index is greater than the end index because the search is done backward.
myIndex = Array.LastIndexOf( myArray, myString, 10, 6 );
Console.WriteLine( "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );

void PrintIndexAndValues( Array anArray )  {
   for ( int i = anArray.GetLowerBound(0); i <= anArray.GetUpperBound(0); i++ )
      Console.WriteLine( "\t[{0}]:\t{1}", i, anArray.GetValue( i ) );
}

/* 
This code produces the following output.

The Array contains the following values:
   [0]:    the
   [1]:    quick
   [2]:    brown
   [3]:    fox
   [4]:    jumps
   [5]:    over
   [6]:    the
   [7]:    lazy
   [8]:    dog
   [9]:    in
   [10]:    the
   [11]:    barn
The last occurrence of "the" is at index 10.
The last occurrence of "the" between the start and index 8 is at index 6.
The last occurrence of "the" between index 5 and index 10 is at index 10.
*/
Public Class SamplesArray    
    
    Public Shared Sub Main()
        
        ' Creates and initializes a new Array with three elements of
        ' the same value.
        Dim myArray As Array = Array.CreateInstance(GetType(String), 12)
        myArray.SetValue("the", 0)
        myArray.SetValue("quick", 1)
        myArray.SetValue("brown", 2)
        myArray.SetValue("fox", 3)
        myArray.SetValue("jumps", 4)
        myArray.SetValue("over", 5)
        myArray.SetValue("the", 6)
        myArray.SetValue("lazy", 7)
        myArray.SetValue("dog", 8)
        myArray.SetValue("in", 9)
        myArray.SetValue("the", 10)
        myArray.SetValue("barn", 11)
        
        ' Displays the values of the Array.
        Console.WriteLine("The Array contains the following values:")
        PrintIndexAndValues(myArray)
        
        ' Searches for the last occurrence of the duplicated value.
        Dim myString As String = "the"
        Dim myIndex As Integer = Array.LastIndexOf(myArray, myString)
        Console.WriteLine("The last occurrence of ""{0}"" is at index {1}.", _
           myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in the first
        ' section of the Array.
        myIndex = Array.LastIndexOf(myArray, myString, 8)
        Console.WriteLine("The last occurrence of ""{0}"" between the start " _
           + "and index 8 is at index {1}.", myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in a section
        ' of the Array.  Note that the start index is greater than the end
        ' index because the search is done backward.
        myIndex = Array.LastIndexOf(myArray, myString, 10, 6)
        Console.WriteLine("The last occurrence of ""{0}"" between index 5 " _
           + "and index 10 is at index {1}.", myString, myIndex)
    End Sub
    
    
    Public Shared Sub PrintIndexAndValues(myArray As Array)
        Dim i As Integer
        For i = myArray.GetLowerBound(0) To myArray.GetUpperBound(0)
            Console.WriteLine(ControlChars.Tab + "[{0}]:" + ControlChars.Tab _
               + "{1}", i, myArray.GetValue(i))
        Next i
    End Sub
End Class

' This code produces the following output.
' 
' The Array contains the following values:
'     [0]:    the
'     [1]:    quick
'     [2]:    brown
'     [3]:    fox
'     [4]:    jumps
'     [5]:    over
'     [6]:    the
'     [7]:    lazy
'     [8]:    dog
'     [9]:    in
'     [10]:    the
'     [11]:    barn
' The last occurrence of "the" is at index 10.
' The last occurrence of "the" between the start and index 8 is at index 6.
' The last occurrence of "the" between index 5 and index 10 is at index 10. 

注解

startIndex 开始向后搜索一维 Array,并在第一个元素处结束。The one-dimensional Array is searched backward starting at startIndex and ending at the first element.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的 Equals 实现。If the element type is a nonintrinsic (user-defined) type, the Equals implementation of that type is used.

由于大多数数组的下限均为零,因此当找不到 value 时,此方法通常会返回-1。Since most arrays will have a lower bound of zero, this method would generally return -1 when value is not found. 在极少数情况下,数组的下限等于 Int32.MinValue 并且找不到 value,则此方法返回 Int32.MaxValue,这 System.Int32.MinValue - 1In the rare case that the lower bound of the array is equal to Int32.MinValue and value is not found, this method returns Int32.MaxValue, which is System.Int32.MinValue - 1.

此方法为 O (n)运算,其中 n 是从 array 开始到 startIndex的元素数。This method is an O(n) operation, where n is the number of elements from the beginning of array to startIndex.

在 .NET Framework 版本2.0 中,此方法使用 ArrayEqualsCompareTo 方法来确定 Object 参数指定的 value 是否存在。In the .NET Framework version 2.0, this method uses the Equals and CompareTo methods of the Array to determine whether the Object specified by the value parameter exists. 在 .NET Framework 的早期版本中,此决定是使用 valueObject 本身的 EqualsCompareTo 方法进行的。In the earlier versions of the .NET Framework, this determination was made by using the Equals and CompareTo methods of the valueObject itself.

另请参阅

LastIndexOf(Array, Object, Int32, Int32)

搜索指定的对象并返回一维 Array 中包含指定数目元素且在指定索引处结尾的元素范围内的最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the one-dimensional Array that contains the specified number of elements and ends at the specified index.

public:
 static int LastIndexOf(Array ^ array, System::Object ^ value, int startIndex, int count);
public static int LastIndexOf (Array array, object value, int startIndex, int count);
static member LastIndexOf : Array * obj * int * int -> int

参数

array
Array

要搜索的一维 ArrayThe one-dimensional Array to search.

value
Object

要在 array 中查找的对象。The object to locate in array.

startIndex
Int32

向后搜索的起始索引。The starting index of the backward search.

count
Int32

要搜索的部分中的元素数。The number of elements in the section to search.

返回

如果在包含 value 中指定的元素数并且在 array 结尾的 count 中的元素范围内找到了 startIndex 的最后一个匹配项的索引,则为该索引;否则为该数组的下限减 1。The index of the last occurrence of value within the range of elements in array that contains the number of elements specified in count and ends at startIndex, if found; otherwise, the lower bound of the array minus 1.

异常

arraynullarray is null.

startIndex 超出了 array 的有效索引范围。startIndex is outside the range of valid indexes for array.

- 或 --or- count 小于零。count is less than zero.

- 或 --or- startIndexcount 未在 array 中指定有效部分。startIndex and count do not specify a valid section in array.

array 是多维的。array is multidimensional.

示例

下面的代码示例演示如何确定指定元素在数组中的最后一个匹配项的索引。The following code example shows how to determine the index of the last occurrence of a specified element in an array. 请注意,LastIndexOf 方法是向后搜索;因此,count 必须小于或等于(startIndex 减去数组的下限加1)。Note that the LastIndexOf method is a backward search; therefore, count must be less than or equal to (startIndex minus the lower bound of the array plus 1).

using namespace System;
void PrintIndexAndValues( Array^ myArray );

void main()
{
   // Creates and initializes a new Array instance with three elements of the same value.
   Array^ myArray = Array::CreateInstance( String::typeid, 12 );
   myArray->SetValue( "the", 0 );
   myArray->SetValue( "quick", 1 );
   myArray->SetValue( "brown", 2 );
   myArray->SetValue( "fox", 3 );
   myArray->SetValue( "jumps", 4 );
   myArray->SetValue( "over", 5 );
   myArray->SetValue( "the", 6 );
   myArray->SetValue( "lazy", 7 );
   myArray->SetValue( "dog", 8 );
   myArray->SetValue( "in", 9 );
   myArray->SetValue( "the", 10 );
   myArray->SetValue( "barn", 11 );

   // Displays the values of the Array.
   Console::WriteLine(  "The Array instance contains the following values:" );
   PrintIndexAndValues( myArray );

   // Searches for the last occurrence of the duplicated value.
   String^ myString =  "the";
   int myIndex = Array::LastIndexOf( myArray, myString );
   Console::WriteLine(  "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in the first section of the Array.
   myIndex = Array::LastIndexOf( myArray, myString, 8 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

   // Searches for the last occurrence of the duplicated value in a section of the Array.  
   // Note that the start index is greater than the end index because the search is done backward.
   myIndex = Array::LastIndexOf( myArray, myString, 10, 6 );
   Console::WriteLine(  "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );
}

void PrintIndexAndValues( Array^ myArray )
{
   for ( int i = myArray->GetLowerBound( 0 ); i <= myArray->GetUpperBound( 0 ); i++ )
      Console::WriteLine(  "\t[{0}]:\t{1}", i, myArray->GetValue( i ) );
}

/* 
 This code produces the following output.
 
 The Array instance contains the following values:
     [0]:    the
     [1]:    quick
     [2]:    brown
     [3]:    fox
     [4]:    jumps
     [5]:    over
     [6]:    the
     [7]:    lazy
     [8]:    dog
     [9]:    in
     [10]:    the
     [11]:    barn
 The last occurrence of "the" is at index 10.
 The last occurrence of "the" between the start and index 8 is at index 6.
 The last occurrence of "the" between index 5 and index 10 is at index 10.
 */
// Creates and initializes a new Array with three elements of the same value.
Array myArray=Array.CreateInstance( typeof(String), 12 );
myArray.SetValue( "the", 0 );
myArray.SetValue( "quick", 1 );
myArray.SetValue( "brown", 2 );
myArray.SetValue( "fox", 3 );
myArray.SetValue( "jumps", 4 );
myArray.SetValue( "over", 5 );
myArray.SetValue( "the", 6 );
myArray.SetValue( "lazy", 7 );
myArray.SetValue( "dog", 8 );
myArray.SetValue( "in", 9 );
myArray.SetValue( "the", 10 );
myArray.SetValue( "barn", 11 );

// Displays the values of the Array.
Console.WriteLine( "The Array contains the following values:" );
PrintIndexAndValues( myArray );

// Searches for the last occurrence of the duplicated value.
String myString = "the";
int myIndex = Array.LastIndexOf( myArray, myString );
Console.WriteLine( "The last occurrence of \"{0}\" is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in the first section of the Array.
myIndex = Array.LastIndexOf( myArray, myString, 8 );
Console.WriteLine( "The last occurrence of \"{0}\" between the start and index 8 is at index {1}.", myString, myIndex );

// Searches for the last occurrence of the duplicated value in a section of the Array.
// Note that the start index is greater than the end index because the search is done backward.
myIndex = Array.LastIndexOf( myArray, myString, 10, 6 );
Console.WriteLine( "The last occurrence of \"{0}\" between index 5 and index 10 is at index {1}.", myString, myIndex );

void PrintIndexAndValues( Array anArray )  {
   for ( int i = anArray.GetLowerBound(0); i <= anArray.GetUpperBound(0); i++ )
      Console.WriteLine( "\t[{0}]:\t{1}", i, anArray.GetValue( i ) );
}

/* 
This code produces the following output.

The Array contains the following values:
   [0]:    the
   [1]:    quick
   [2]:    brown
   [3]:    fox
   [4]:    jumps
   [5]:    over
   [6]:    the
   [7]:    lazy
   [8]:    dog
   [9]:    in
   [10]:    the
   [11]:    barn
The last occurrence of "the" is at index 10.
The last occurrence of "the" between the start and index 8 is at index 6.
The last occurrence of "the" between index 5 and index 10 is at index 10.
*/
Public Class SamplesArray    
    
    Public Shared Sub Main()
        
        ' Creates and initializes a new Array with three elements of
        ' the same value.
        Dim myArray As Array = Array.CreateInstance(GetType(String), 12)
        myArray.SetValue("the", 0)
        myArray.SetValue("quick", 1)
        myArray.SetValue("brown", 2)
        myArray.SetValue("fox", 3)
        myArray.SetValue("jumps", 4)
        myArray.SetValue("over", 5)
        myArray.SetValue("the", 6)
        myArray.SetValue("lazy", 7)
        myArray.SetValue("dog", 8)
        myArray.SetValue("in", 9)
        myArray.SetValue("the", 10)
        myArray.SetValue("barn", 11)
        
        ' Displays the values of the Array.
        Console.WriteLine("The Array contains the following values:")
        PrintIndexAndValues(myArray)
        
        ' Searches for the last occurrence of the duplicated value.
        Dim myString As String = "the"
        Dim myIndex As Integer = Array.LastIndexOf(myArray, myString)
        Console.WriteLine("The last occurrence of ""{0}"" is at index {1}.", _
           myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in the first
        ' section of the Array.
        myIndex = Array.LastIndexOf(myArray, myString, 8)
        Console.WriteLine("The last occurrence of ""{0}"" between the start " _
           + "and index 8 is at index {1}.", myString, myIndex)
        
        ' Searches for the last occurrence of the duplicated value in a section
        ' of the Array.  Note that the start index is greater than the end
        ' index because the search is done backward.
        myIndex = Array.LastIndexOf(myArray, myString, 10, 6)
        Console.WriteLine("The last occurrence of ""{0}"" between index 5 " _
           + "and index 10 is at index {1}.", myString, myIndex)
    End Sub
    
    
    Public Shared Sub PrintIndexAndValues(myArray As Array)
        Dim i As Integer
        For i = myArray.GetLowerBound(0) To myArray.GetUpperBound(0)
            Console.WriteLine(ControlChars.Tab + "[{0}]:" + ControlChars.Tab _
               + "{1}", i, myArray.GetValue(i))
        Next i
    End Sub
End Class

' This code produces the following output.
' 
' The Array contains the following values:
'     [0]:    the
'     [1]:    quick
'     [2]:    brown
'     [3]:    fox
'     [4]:    jumps
'     [5]:    over
'     [6]:    the
'     [7]:    lazy
'     [8]:    dog
'     [9]:    in
'     [10]:    the
'     [11]:    barn
' The last occurrence of "the" is at index 10.
' The last occurrence of "the" between the start and index 8 is at index 6.
' The last occurrence of "the" between index 5 and index 10 is at index 10. 

注解

startIndex 开始向后搜索一维 Array,在 startIndex 减号处结束,如果 count 大于0,则从 count 减号结束。The one-dimensional Array is searched backward starting at startIndex and ending at startIndex minus count plus 1, if count is greater than 0.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的Equals 实现。If the element type is a nonintrinsic (user-defined) type, theEquals implementation of that type is used.

由于大多数数组的下限均为零,因此当找不到 value 时,此方法通常会返回-1。Since most arrays will have a lower bound of zero, this method would generally return -1 when value is not found. 在极少数情况下,数组的下限等于 Int32.MinValue 并且找不到 value,则此方法返回 Int32.MaxValue,这 System.Int32.MinValue - 1In the rare case that the lower bound of the array is equal to Int32.MinValue and value is not found, this method returns Int32.MaxValue, which is System.Int32.MinValue - 1.

此方法是一个 O (n)操作,其中 ncount的。This method is an O(n) operation, where n is count.

在 .NET Framework 版本2.0 中,此方法使用 ArrayEqualsCompareTo 方法来确定 Object 参数指定的 value 是否存在。In the .NET Framework version 2.0, this method uses the Equals and CompareTo methods of the Array to determine whether the Object specified by the value parameter exists. 在 .NET Framework 的早期版本中,此决定是使用 valueObject 本身的 EqualsCompareTo 方法进行的。In the earlier versions of the .NET Framework, this determination was made by using the Equals and CompareTo methods of the valueObject itself.

另请参阅

LastIndexOf<T>(T[], T)

搜索指定的对象,并返回整个 Array 中最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the entire Array.

public:
generic <typename T>
 static int LastIndexOf(cli::array <T> ^ array, T value);
public static int LastIndexOf<T> (T[] array, T value);
static member LastIndexOf : 'T[] * 'T -> int
Public Shared Function LastIndexOf(Of T) (array As T(), value As T) As Integer

类型参数

T

数组元素的类型。The type of the elements of the array.

参数

array
T[]

要搜索的从零开始的一维 ArrayThe one-dimensional, zero-based Array to search.

value
T

要在 array 中查找的对象。The object to locate in array.

返回

如果在整个 value 中找到 array 的最后一个匹配项从零开始的索引,则为该索引;否则为 -1。The zero-based index of the last occurrence of value within the entire array, if found; otherwise, -1.

异常

arraynullarray is null.

示例

下面的代码示例演示 LastIndexOf 方法的所有三个泛型重载。The following code example demonstrates all three generic overloads of the LastIndexOf method. 创建一个字符串数组,其中有一个条目出现在索引位置0和索引位置5。An array of strings is created, with one entry that appears twice, at index location 0 and index location 5. LastIndexOf<T>(T[], T) 方法重载从末尾搜索整个数组,并查找字符串的第二个匹配项。The LastIndexOf<T>(T[], T) method overload searches the entire array from the end, and finds the second occurrence of the string. LastIndexOf<T>(T[], T, Int32) 方法重载用于从索引位置3开始向后搜索数组,并继续到数组的开头,并查找字符串的第一个匹配项。The LastIndexOf<T>(T[], T, Int32) method overload is used to search the array backward beginning with index location 3 and continuing to the beginning of the array, and finds the first occurrence of the string. 最后,LastIndexOf<T>(T[], T, Int32, Int32) 方法重载用于搜索由四个条目组成的范围,从索引位置4开始并向后扩展(也就是说,它会搜索位于4、3、2和1位置的项);此搜索将返回-1,因为该范围中没有搜索字符串的实例。Finally, the LastIndexOf<T>(T[], T, Int32, Int32) method overload is used to search a range of four entries, beginning at index location 4 and extending backward (that is, it searches the items at locations 4, 3, 2, and 1); this search returns -1 because there are no instances of the search string in that range.

using namespace System;

void main()
{
    array<String^>^ dinosaurs = { "Tyrannosaurus", 
        "Amargasaurus",
        "Mamenchisaurus",
        "Brachiosaurus",
        "Deinonychus",
        "Tyrannosaurus",
        "Compsognathus" };

    Console::WriteLine();
    for each(String^ dinosaur in dinosaurs )
    {
        Console::WriteLine(dinosaur);
    }

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus"));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));
}

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
 */
string[] dinosaurs = { "Tyrannosaurus",
    "Amargasaurus",
    "Mamenchisaurus",
    "Brachiosaurus",
    "Deinonychus",
    "Tyrannosaurus",
    "Compsognathus" };

Console.WriteLine();
foreach(string dinosaur in dinosaurs)
{
    Console.WriteLine(dinosaur);
}

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus"));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
*/
Public Class Example

    Public Shared Sub Main()

        Dim dinosaurs() As String = { "Tyrannosaurus", _
            "Amargasaurus", _
            "Mamenchisaurus", _
            "Brachiosaurus", _
            "Deinonychus", _
            "Tyrannosaurus", _
            "Compsognathus" }

        Console.WriteLine()
        For Each dinosaur As String In dinosaurs
            Console.WriteLine(dinosaur)
        Next

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus""): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus"))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 3): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 4, 4): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4))

    End Sub
End Class

' This code example produces the following output:
'
'Tyrannosaurus
'Amargasaurus
'Mamenchisaurus
'Brachiosaurus
'Deinonychus
'Tyrannosaurus
'Compsognathus
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1

注解

从最后一个元素开始向后搜索 Array,并在第一个元素处结束。The Array is searched backward starting at the last element and ending at the first element.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的 Equals 实现。If the element type is a nonintrinsic (user-defined) type, the Equals implementation of that type is used.

此方法是一个 O (n)操作,其中 narrayLengthThis method is an O(n) operation, where n is the Length of array.

另请参阅

LastIndexOf<T>(T[], T, Int32)

搜索指定的对象并返回 Array 中从第一个元素到指定索引的元素范围内最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the Array that extends from the first element to the specified index.

public:
generic <typename T>
 static int LastIndexOf(cli::array <T> ^ array, T value, int startIndex);
public static int LastIndexOf<T> (T[] array, T value, int startIndex);
static member LastIndexOf : 'T[] * 'T * int -> int
Public Shared Function LastIndexOf(Of T) (array As T(), value As T, startIndex As Integer) As Integer

类型参数

T

数组元素的类型。The type of the elements of the array.

参数

array
T[]

要搜索的从零开始的一维 ArrayThe one-dimensional, zero-based Array to search.

value
T

要在 array 中查找的对象。The object to locate in array.

startIndex
Int32

向后搜索的从零开始的起始索引。The zero-based starting index of the backward search.

返回

如果在 value 中从第一个元素到 array 的元素范围中找到了该 startIndex 最后一个匹配项从零开始的索引,则为该索引;否则为 -1。The zero-based index of the last occurrence of value within the range of elements in array that extends from the first element to startIndex, if found; otherwise, -1.

异常

arraynullarray is null.

startIndex 超出了 array 的有效索引范围。startIndex is outside the range of valid indexes for array.

示例

下面的代码示例演示 LastIndexOf 方法的所有三个泛型重载。The following code example demonstrates all three generic overloads of the LastIndexOf method. 创建一个字符串数组,其中有一个条目出现在索引位置0和索引位置5。An array of strings is created, with one entry that appears twice, at index location 0 and index location 5. LastIndexOf<T>(T[], T) 方法重载从末尾搜索整个数组,并查找字符串的第二个匹配项。The LastIndexOf<T>(T[], T) method overload searches the entire array from the end, and finds the second occurrence of the string. LastIndexOf<T>(T[], T, Int32) 方法重载用于从索引位置3开始向后搜索数组,并继续到数组的开头,并查找字符串的第一个匹配项。The LastIndexOf<T>(T[], T, Int32) method overload is used to search the array backward beginning with index location 3 and continuing to the beginning of the array, and finds the first occurrence of the string. 最后,LastIndexOf<T>(T[], T, Int32, Int32) 方法重载用于搜索由四个条目组成的范围,从索引位置4开始并向后扩展(也就是说,它会搜索位于4、3、2和1位置的项);此搜索将返回-1,因为该范围中没有搜索字符串的实例。Finally, the LastIndexOf<T>(T[], T, Int32, Int32) method overload is used to search a range of four entries, beginning at index location 4 and extending backward (that is, it searches the items at locations 4, 3, 2, and 1); this search returns -1 because there are no instances of the search string in that range.

using namespace System;

void main()
{
    array<String^>^ dinosaurs = { "Tyrannosaurus", 
        "Amargasaurus",
        "Mamenchisaurus",
        "Brachiosaurus",
        "Deinonychus",
        "Tyrannosaurus",
        "Compsognathus" };

    Console::WriteLine();
    for each(String^ dinosaur in dinosaurs )
    {
        Console::WriteLine(dinosaur);
    }

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus"));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));
}

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
 */
string[] dinosaurs = { "Tyrannosaurus",
    "Amargasaurus",
    "Mamenchisaurus",
    "Brachiosaurus",
    "Deinonychus",
    "Tyrannosaurus",
    "Compsognathus" };

Console.WriteLine();
foreach(string dinosaur in dinosaurs)
{
    Console.WriteLine(dinosaur);
}

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus"));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
*/
Public Class Example

    Public Shared Sub Main()

        Dim dinosaurs() As String = { "Tyrannosaurus", _
            "Amargasaurus", _
            "Mamenchisaurus", _
            "Brachiosaurus", _
            "Deinonychus", _
            "Tyrannosaurus", _
            "Compsognathus" }

        Console.WriteLine()
        For Each dinosaur As String In dinosaurs
            Console.WriteLine(dinosaur)
        Next

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus""): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus"))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 3): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 4, 4): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4))

    End Sub
End Class

' This code example produces the following output:
'
'Tyrannosaurus
'Amargasaurus
'Mamenchisaurus
'Brachiosaurus
'Deinonychus
'Tyrannosaurus
'Compsognathus
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1

注解

startIndex 开始向后搜索 Array,并在第一个元素处结束。The Array is searched backward starting at startIndex and ending at the first element.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的 Equals 实现。If the element type is a nonintrinsic (user-defined) type, the Equals implementation of that type is used.

此方法为 O (n)运算,其中 n 是从 array 开始到 startIndex的元素数。This method is an O(n) operation, where n is the number of elements from the beginning of array to startIndex.

另请参阅

LastIndexOf<T>(T[], T, Int32, Int32)

搜索指定的对象,并返回 Array 中包含指定数目元素并在指定索引处结尾的元素范围内的最后一个匹配项的索引。Searches for the specified object and returns the index of the last occurrence within the range of elements in the Array that contains the specified number of elements and ends at the specified index.

public:
generic <typename T>
 static int LastIndexOf(cli::array <T> ^ array, T value, int startIndex, int count);
public static int LastIndexOf<T> (T[] array, T value, int startIndex, int count);
static member LastIndexOf : 'T[] * 'T * int * int -> int
Public Shared Function LastIndexOf(Of T) (array As T(), value As T, startIndex As Integer, count As Integer) As Integer

类型参数

T

数组元素的类型。The type of the elements of the array.

参数

array
T[]

要搜索的从零开始的一维 ArrayThe one-dimensional, zero-based Array to search.

value
T

要在 array 中查找的对象。The object to locate in array.

startIndex
Int32

向后搜索的从零开始的起始索引。The zero-based starting index of the backward search.

count
Int32

要搜索的部分中的元素数。The number of elements in the section to search.

返回

如果在 value 中找到了包含 array 中指定元素数并且到 count 结尾的元素范围内的 startIndex 的最后一个匹配项的从零开始的索引,则为该索引;否则为 -1。The zero-based index of the last occurrence of value within the range of elements in array that contains the number of elements specified in count and ends at startIndex, if found; otherwise, -1.

异常

arraynullarray is null.

startIndex 超出了 array 的有效索引范围。startIndex is outside the range of valid indexes for array.

- 或 --or- count 小于零。count is less than zero.

- 或 --or- startIndexcount 未在 array 中指定有效部分。startIndex and count do not specify a valid section in array.

示例

下面的代码示例演示 LastIndexOf 方法的所有三个泛型重载。The following code example demonstrates all three generic overloads of the LastIndexOf method. 创建一个字符串数组,其中有一个条目出现在索引位置0和索引位置5。An array of strings is created, with one entry that appears twice, at index location 0 and index location 5. LastIndexOf<T>(T[], T) 方法重载从末尾搜索整个数组,并查找字符串的第二个匹配项。The LastIndexOf<T>(T[], T) method overload searches the entire array from the end, and finds the second occurrence of the string. LastIndexOf<T>(T[], T, Int32) 方法重载用于从索引位置3开始向后搜索数组,并继续到数组的开头,并查找字符串的第一个匹配项。The LastIndexOf<T>(T[], T, Int32) method overload is used to search the array backward beginning with index location 3 and continuing to the beginning of the array, and finds the first occurrence of the string. 最后,LastIndexOf<T>(T[], T, Int32, Int32) 方法重载用于搜索由四个条目组成的范围,从索引位置4开始并向后扩展(也就是说,它会搜索位于4、3、2和1位置的项);此搜索将返回-1,因为该范围中没有搜索字符串的实例。Finally, the LastIndexOf<T>(T[], T, Int32, Int32) method overload is used to search a range of four entries, beginning at index location 4 and extending backward (that is, it searches the items at locations 4, 3, 2, and 1); this search returns -1 because there are no instances of the search string in that range.

using namespace System;

void main()
{
    array<String^>^ dinosaurs = { "Tyrannosaurus", 
        "Amargasaurus",
        "Mamenchisaurus",
        "Brachiosaurus",
        "Deinonychus",
        "Tyrannosaurus",
        "Compsognathus" };

    Console::WriteLine();
    for each(String^ dinosaur in dinosaurs )
    {
        Console::WriteLine(dinosaur);
    }

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus"));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

    Console::WriteLine(
        "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
        Array::LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));
}

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
 */
string[] dinosaurs = { "Tyrannosaurus",
    "Amargasaurus",
    "Mamenchisaurus",
    "Brachiosaurus",
    "Deinonychus",
    "Tyrannosaurus",
    "Compsognathus" };

Console.WriteLine();
foreach(string dinosaur in dinosaurs)
{
    Console.WriteLine(dinosaur);
}

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus"));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3));

Console.WriteLine(
    "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): {0}", 
    Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4));

/* This code example produces the following output:

Tyrannosaurus
Amargasaurus
Mamenchisaurus
Brachiosaurus
Deinonychus
Tyrannosaurus
Compsognathus

Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1
*/
Public Class Example

    Public Shared Sub Main()

        Dim dinosaurs() As String = { "Tyrannosaurus", _
            "Amargasaurus", _
            "Mamenchisaurus", _
            "Brachiosaurus", _
            "Deinonychus", _
            "Tyrannosaurus", _
            "Compsognathus" }

        Console.WriteLine()
        For Each dinosaur As String In dinosaurs
            Console.WriteLine(dinosaur)
        Next

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus""): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus"))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 3): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3))

        Console.WriteLine(vbLf & _
            "Array.LastIndexOf(dinosaurs, ""Tyrannosaurus"", 4, 4): {0}", _
            Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4))

    End Sub
End Class

' This code example produces the following output:
'
'Tyrannosaurus
'Amargasaurus
'Mamenchisaurus
'Brachiosaurus
'Deinonychus
'Tyrannosaurus
'Compsognathus
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus"): 5
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3): 0
'
'Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4): -1

注解

将从 startIndex 开始向后 Array 搜索,如果 count 大于0,则从 startIndex 减号开始,在 count 减号处结束。The Array is searched backward starting at startIndex and ending at startIndex minus count plus 1, if count is greater than 0.

使用 Object.Equals 方法将元素与指定的值进行比较。The elements are compared to the specified value using the Object.Equals method. 如果元素类型为 nonintrinsic (用户定义)类型,则使用该类型的 Equals 实现。If the element type is a nonintrinsic (user-defined) type, the Equals implementation of that type is used.

此方法是一个 O (n)操作,其中 ncount的。This method is an O(n) operation, where n is count.

另请参阅

适用于