Enumerable.Append<TSource>(IEnumerable<TSource>, TSource) メソッド

定義

シーケンスの末尾に値を追加します。

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

型パラメーター

TSource

source の要素の型。

パラメーター

source
IEnumerable<TSource>

値のシーケンス。

element
TSource

source に追加する値。

戻り値

IEnumerable<TSource>

element で終わる新しいシーケンス。

例外

sourcenullです。

次のコード例では、 を使用 Append してシーケンスの末尾に値を追加する方法を示します。

// Creating a list of numbers
List<int> numbers = new List<int> { 1, 2, 3, 4 };

// Trying to append any value of the same type
numbers.Append(5);

// It doesn't work because the original list has not been changed
Console.WriteLine(string.Join(", ", numbers));

// It works now because we are using a changed copy of the original list
Console.WriteLine(string.Join(", ", numbers.Append(5)));

// If you prefer, you can create a new list explicitly
List<int> newNumbers = numbers.Append(5).ToList();

// And then write to the console output
Console.WriteLine(string.Join(", ", newNumbers));

// This code produces the following output:
//
// 1, 2, 3, 4
// 1, 2, 3, 4, 5
// 1, 2, 3, 4, 5
' Creating a list of numbers
Dim numbers As New List(Of Integer)(New Integer() {1, 2, 3, 4})

' Trying to append any value of the same type
numbers.Append(5)

' It doesn't work because the original list has not been changed
Console.WriteLine(String.Join(", ", numbers))

' It works now because we are using a changed copy of the original list
Console.WriteLine(String.Join(", ", numbers.Append(5)))

' If you prefer, you can create a new list explicitly
Dim newNumbers As List(Of Integer) = numbers.Append(5).ToList

' And then write to the console output
Console.WriteLine(String.Join(", ", newNumbers))

' This code produces the following output:
'
' 1, 2, 3, 4
' 1, 2, 3, 4, 5
' 1, 2, 3, 4, 5

注釈

Note

このメソッドは、コレクションの要素を変更しません。 代わりに、新しい要素を使用してコレクションのコピーを作成します。

適用対象