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Enumerable.Append<TSource>(IEnumerable<TSource>, TSource) Método

Definición

Anexa un valor al final de la secuencia.

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)

Parámetros de tipo

TSource

Tipo de los elementos de source.

Parámetros

source
IEnumerable<TSource>

Secuencia de valores.

element
TSource

Valor que se va a anexar a source.

Devoluciones

IEnumerable<TSource>

Nueva secuencia que termina en element.

Excepciones

source es null.

Ejemplos

En el ejemplo de código siguiente se muestra cómo usar Append para anexar un valor al final de la secuencia.

// 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

Comentarios

Nota

Este método no modifica los elementos de la colección. En su lugar, crea una copia de la colección con el nuevo elemento .

Se aplica a