语言

Array.LastIndexOf 方法

定义

返回一维 Array 或部分值的最后一个匹配项的 Array索引。

重载

名称 说明
LastIndexOf(Array, Object)

搜索指定的对象,并返回整个一维 Array内最后一个匹配项的索引。

LastIndexOf(Array, Object, Int32)

搜索指定的对象,并在从第一个元素扩展到指定索引的一维 Array 元素范围内返回最后一个匹配项的索引。

LastIndexOf(Array, Object, Int32, Int32)

搜索指定的对象,并返回一维 Array 元素范围内最后一个匹配项的索引,该元素包含指定数量的元素,并在指定索引处结束。

LastIndexOf<T>(T[], T)

搜索指定的对象并返回整个 Array内最后一个匹配项的索引。

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

搜索指定的对象,并在从第一个元素扩展到指定索引的元素 Array 范围内返回最后一个匹配项的索引。

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

搜索指定的对象,并返回包含指定数量的元素并在指定索引处结束的元素 Array 范围内的最后一个匹配项的索引。

LastIndexOf(Array, Object)

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象,并返回整个一维 Array内最后一个匹配项的索引。

public:
 static int LastIndexOf(Array ^ array, System::Object ^ value);
public static int LastIndexOf(Array array, object value);
public static int LastIndexOf(Array array, object? value);
static member LastIndexOf : Array * obj -> int
Public Shared Function LastIndexOf (array As Array, value As Object) As Integer

参数

array
Array

要搜索的一维 Array 。

value
Object

要查找到 array的对象的对象。

返回

如果找到,则为整个value中最后一个匹配项array的索引;否则,数组的下限减 1。

例外

array 是 null。

array 是多维。

示例

下面的代码示例演示如何确定数组中指定元素的最后一个匹配项的索引。

let printIndexAndValues (arr: 'a []) =
   for i = arr.GetLowerBound 0 to arr.GetUpperBound 0 do
      printfn $"\t[{i}]:\t{arr[i]}"

// Creates and initializes a new Array with three elements of the same value.
let myArray = 
   [| "the"; "quick"; "brown"; "fox"
      "jumps"; "over"; "the"; "lazy"
      "dog"; "in"; "the"; "barn" |]

// Displays the values of the Array.
printfn "The Array contains the following values:"
printIndexAndValues myArray

// Searches for the last occurrence of the duplicated value.
let myString = "the"
let myIndex = Array.LastIndexOf(myArray, myString)
printfn $"The last occurrence of \"{myString}\" is at index {myIndex}."

// Searches for the last occurrence of the duplicated value in the first section of the Array.
let myIndex = Array.LastIndexOf(myArray, myString, 8)
printfn $"The last occurrence of \"{myString}\" between the start and index 8 is at index {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.
let myIndex = Array.LastIndexOf( myArray, myString, 10, 6 )
printfn $"The last occurrence of \"{myString}\" between index 5 and index 10 is at index {myIndex}."


//      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.
// 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

一维 Array 从最后一个元素向后搜索,最后一个元素结束。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型, Equals 则使用该类型的实现。

由于大多数数组的下限为零,因此此方法通常在找不到时 value 返回 -1。 在极少数情况下,数组的下限等于Int32.MinValuevalue和未找到,此方法返回Int32.MaxValue,即System.Int32.MinValue - 1。

此方法是 O(n) 操作,其位置 n 为 Lengtharray.

此方法使用Equals该方法和CompareTo方法Array来确定参数指定的Object是否存在value。

CompareTo item集合中对象上参数的方法。

另请参阅

适用于

LastIndexOf(Array, Object, Int32)

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象,并在从第一个元素扩展到指定索引的一维 Array 元素范围内返回最后一个匹配项的索引。

public:
 static int LastIndexOf(Array ^ array, System::Object ^ value, int startIndex);
public static int LastIndexOf(Array array, object value, int startIndex);
public static int LastIndexOf(Array array, object? value, int startIndex);
static member LastIndexOf : Array * obj * int -> int
Public Shared Function LastIndexOf (array As Array, value As Object, startIndex As Integer) As Integer

参数

array
Array

要搜索的一维 Array 。

value
Object

要查找到 array的对象的对象。

startIndex
Int32

向后搜索的起始索引。

返回

从第一个元素扩展到第一个元素valuearray范围内的最后一个匹配startIndex项的索引(如果找到);否则,数组的下限减 1。

例外

array 是 null。

startIndex 超出了有效索引 array的范围。

array 是多维。

示例

下面的代码示例演示如何确定数组中指定元素的最后一个匹配项的索引。

let printIndexAndValues (arr: 'a []) =
   for i = arr.GetLowerBound 0 to arr.GetUpperBound 0 do
      printfn $"\t[{i}]:\t{arr[i]}"

// Creates and initializes a new Array with three elements of the same value.
let myArray = 
   [| "the"; "quick"; "brown"; "fox"
      "jumps"; "over"; "the"; "lazy"
      "dog"; "in"; "the"; "barn" |]

// Displays the values of the Array.
printfn "The Array contains the following values:"
printIndexAndValues myArray

// Searches for the last occurrence of the duplicated value.
let myString = "the"
let myIndex = Array.LastIndexOf(myArray, myString)
printfn $"The last occurrence of \"{myString}\" is at index {myIndex}."

// Searches for the last occurrence of the duplicated value in the first section of the Array.
let myIndex = Array.LastIndexOf(myArray, myString, 8)
printfn $"The last occurrence of \"{myString}\" between the start and index 8 is at index {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.
let myIndex = Array.LastIndexOf( myArray, myString, 10, 6 )
printfn $"The last occurrence of \"{myString}\" between index 5 and index 10 is at index {myIndex}."


//      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.
// 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

从第一个元素开始Array和结束搜索一维startIndex。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型, Equals 则使用该类型的实现。

由于大多数数组的下限为零,因此此方法通常在找不到时 value 返回 -1。 在极少数情况下,数组的下限等于Int32.MinValuevalue和未找到,此方法返回Int32.MaxValue,即System.Int32.MinValue - 1。

此方法是一个 O(n) 操作,其中 n 元素数从开头 array 到 startIndex。

此方法使用Equals该方法和CompareTo方法Array来确定参数指定的Object是否存在value。

另请参阅

适用于

LastIndexOf(Array, Object, Int32, Int32)

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象,并返回一维 Array 元素范围内最后一个匹配项的索引,该元素包含指定数量的元素,并在指定索引处结束。

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);
public static int LastIndexOf(Array array, object? value, int startIndex, int count);
static member LastIndexOf : Array * obj * int * int -> int
Public Shared Function LastIndexOf (array As Array, value As Object, startIndex As Integer, count As Integer) As Integer

参数

array
Array

要搜索的一维 Array 。

value
Object

要查找到 array的对象的对象。

startIndex
Int32

向后搜索的起始索引。

count
Int32

要搜索的节中的元素数。

返回

如果找到数组的下限减 1,则为元素范围中value最后一个匹配array项的索引,其中包含指定count元素的数目和结尾startIndex处。否则,数组的下限为减 1。

例外

array 是 null。

startIndex 超出了有效索引 array的范围。

-或-

count 小于零。

-或-

startIndex 且 count 未在 . 中 array指定有效节。

array 是多维。

示例

下面的代码示例演示如何确定数组中指定元素的最后一个匹配项的索引。 请注意, LastIndexOf 该方法是向后搜索;因此, count 必须小于或等于(startIndex 减去数组的下限加 1)。

let printIndexAndValues (arr: 'a []) =
   for i = arr.GetLowerBound 0 to arr.GetUpperBound 0 do
      printfn $"\t[{i}]:\t{arr[i]}"

// Creates and initializes a new Array with three elements of the same value.
let myArray = 
   [| "the"; "quick"; "brown"; "fox"
      "jumps"; "over"; "the"; "lazy"
      "dog"; "in"; "the"; "barn" |]

// Displays the values of the Array.
printfn "The Array contains the following values:"
printIndexAndValues myArray

// Searches for the last occurrence of the duplicated value.
let myString = "the"
let myIndex = Array.LastIndexOf(myArray, myString)
printfn $"The last occurrence of \"{myString}\" is at index {myIndex}."

// Searches for the last occurrence of the duplicated value in the first section of the Array.
let myIndex = Array.LastIndexOf(myArray, myString, 8)
printfn $"The last occurrence of \"{myString}\" between the start and index 8 is at index {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.
let myIndex = Array.LastIndexOf( myArray, myString, 10, 6 )
printfn $"The last occurrence of \"{myString}\" between index 5 and index 10 is at index {myIndex}."


//      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.
// 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

如果大于 0,Array则从负startIndex加 1 开始startIndexcount向后搜索一维count。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型,Equals 则使用该类型的实现。

由于大多数数组的下限为零,因此此方法通常在找不到时 value 返回 -1。 在极少数情况下,数组的下限等于Int32.MinValuevalue和未找到,此方法返回Int32.MaxValue,即System.Int32.MinValue - 1。

此方法是 O(n) 操作,其中 n 。count

此方法使用Equals该方法和CompareTo方法Array来确定参数指定的Object是否存在value。

另请参阅

适用于

LastIndexOf<T>(T[], T)

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象并返回整个 Array内最后一个匹配项的索引。

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。

参数

array
T[]

从零开始 Array 的一维搜索。

value
T

要查找到 array的对象的对象。

返回

如果找到,则为整个value中最后一个匹配项array的从零开始的索引;否则为 -1。

例外

array 是 null。

示例

下面的代码示例演示了该方法的所有三个 LastIndexOf 泛型重载。 创建字符串数组,其中一个条目出现在索引位置 0 和索引位置 5 处两次。 该方法 LastIndexOf<T>(T[], T) 重载从末尾搜索整个数组,并查找字符串的第二个匹配项。 方法 LastIndexOf<T>(T[], T, Int32) 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。 最后,该方法 LastIndexOf<T>(T[], T, Int32, Int32) 重载用于搜索从索引位置 4 开始的四个条目范围并向后扩展(也就是说,它会在位置 4、3、2 和 1 处搜索项);此搜索返回 -1,因为该区域中没有搜索字符串的实例。

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
*/
open System

let dinosaurs = 
    [| "Tyrannosaurus"
       "Amargasaurus"
       "Mamenchisaurus"
       "Brachiosaurus"
       "Deinonychus"
       "Tyrannosaurus"
       "Compsognathus" |]

printfn ""
for dino in dinosaurs do
    printfn $"{dino}"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus")
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): %i"

// 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 Array 最后一个元素开始搜索向后搜索,最后一个元素结束。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型, Equals 则使用该类型的实现。

此方法是 O(n) 操作,其位置 n 为 Lengtharray.

另请参阅

适用于

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

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象,并在从第一个元素扩展到指定索引的元素 Array 范围内返回最后一个匹配项的索引。

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。

参数

array
T[]

从零开始 Array 的一维搜索。

value
T

要查找到 array的对象的对象。

startIndex
Int32

从零开始的向后搜索索引。

返回

从第一个元素扩展到第一个元素valuearray范围内的最后一个匹配startIndex项的从零开始的索引(如果找到);否则为 -1。

例外

array 是 null。

startIndex 超出了有效索引 array的范围。

示例

下面的代码示例演示了该方法的所有三个 LastIndexOf 泛型重载。 创建字符串数组,其中一个条目出现在索引位置 0 和索引位置 5 处两次。 该方法 LastIndexOf<T>(T[], T) 重载从末尾搜索整个数组,并查找字符串的第二个匹配项。 方法 LastIndexOf<T>(T[], T, Int32) 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。 最后,该方法 LastIndexOf<T>(T[], T, Int32, Int32) 重载用于搜索从索引位置 4 开始的四个条目范围并向后扩展(也就是说,它会在位置 4、3、2 和 1 处搜索项);此搜索返回 -1,因为该区域中没有搜索字符串的实例。

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
*/
open System

let dinosaurs = 
    [| "Tyrannosaurus"
       "Amargasaurus"
       "Mamenchisaurus"
       "Brachiosaurus"
       "Deinonychus"
       "Tyrannosaurus"
       "Compsognathus" |]

printfn ""
for dino in dinosaurs do
    printfn $"{dino}"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus")
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): %i"

// 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 Array 第一个元素开始 startIndex 和结束搜索后移。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型, Equals 则使用该类型的实现。

此方法是一个 O(n) 操作,其中 n 元素数从开头 array 到 startIndex。

另请参阅

适用于

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

Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs
Source:
Array.cs

搜索指定的对象,并返回包含指定数量的元素并在指定索引处结束的元素 Array 范围内的最后一个匹配项的索引。

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。

参数

array
T[]

从零开始 Array 的一维搜索。

value
T

要查找到 array的对象的对象。

startIndex
Int32

从零开始的向后搜索索引。

count
Int32

要搜索的节中的元素数。

返回

在元素范围内最后一个匹配 value 项的从零开始的 array 索引,该索引包含指定的 count 元素数和结尾 startIndex处(如果找到);否则为 -1。

例外

array 是 null。

startIndex 超出了有效索引 array的范围。

-或-

count 小于零。

-或-

startIndex 且 count 未在 . 中 array指定有效节。

示例

下面的代码示例演示了该方法的所有三个 LastIndexOf 泛型重载。 创建字符串数组,其中一个条目出现在索引位置 0 和索引位置 5 处两次。 该方法 LastIndexOf<T>(T[], T) 重载从末尾搜索整个数组,并查找字符串的第二个匹配项。 方法 LastIndexOf<T>(T[], T, Int32) 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。 最后,该方法 LastIndexOf<T>(T[], T, Int32, Int32) 重载用于搜索从索引位置 4 开始的四个条目范围并向后扩展(也就是说,它会在位置 4、3、2 和 1 处搜索项);此搜索返回 -1,因为该区域中没有搜索字符串的实例。

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
*/
open System

let dinosaurs = 
    [| "Tyrannosaurus"
       "Amargasaurus"
       "Mamenchisaurus"
       "Brachiosaurus"
       "Deinonychus"
       "Tyrannosaurus"
       "Compsognathus" |]

printfn ""
for dino in dinosaurs do
    printfn $"{dino}"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus")
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\"): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 3)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 3): %i"

Array.LastIndexOf(dinosaurs, "Tyrannosaurus", 4, 4)
|> printfn "\nArray.LastIndexOf(dinosaurs, \"Tyrannosaurus\", 4, 4): %i"

// 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

如果Array大于 0,startIndex则从负startIndex加 1 开始搜索count后向count搜索。

使用该方法将元素与指定的值 Object.Equals 进行比较。 如果元素类型是非内联类型(用户定义的)类型, Equals 则使用该类型的实现。

此方法是 O(n) 操作,其中 n 。count

另请参阅

适用于