Enumerable.Prepend<TSource>(IEnumerable<TSource>, TSource) Méthode

Définition

Ajoute une valeur au début de la séquence.Adds a value to the beginning of the sequence.

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

Paramètres de type

TSource

Le type des éléments de source.The type of the elements of source.

Paramètres

source
IEnumerable<TSource>

Séquence de valeurs.A sequence of values.

element
TSource

Valeur à ajouter à source.The value to prepend to source.

Retours

IEnumerable<TSource>

Nouvelle séquence qui commence par element.A new sequence that begins with element.

Exceptions

source a la valeur null.source is null.

Exemples

L’exemple de code suivant montre comment utiliser Prepend pour ajouter une valeur au début de la séquence.The following code example demonstrates how to use Prepend to prepend a value to the beginning of the sequence.

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

// Trying to prepend any value of the same type
numbers.Prepend(0);

// 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.Prepend(0)));

// If you prefer, you can create a new list explicitly
List<int> newNumbers = numbers.Prepend(0).ToList();

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

// This code produces the following output:
//
// 1, 2, 3, 4
// 0, 1, 2, 3, 4
// 0, 1, 2, 3, 4
' Creating a list of numbers
Dim numbers As New List(Of Integer)(New Integer() {1, 2, 3, 4})

' Trying to prepend any value of the same type
numbers.Prepend(0)

' 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.Prepend(0)))

' If you prefer, you can create a new list explicitly
Dim newNumbers As List(Of Integer) = numbers.Prepend(0).ToList

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

' This code produces the following output:
'
' 1, 2, 3, 4
' 0, 1, 2, 3, 4
' 0, 1, 2, 3, 4

Remarques

Notes

Cette méthode ne modifie pas les éléments de la collection.This method does not modify the elements of the collection. Au lieu de cela, il crée une copie de la collection avec le nouvel élément.Instead, it creates a copy of the collection with the new element.

S’applique à