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

Definition

Fügt einen Wert am Anfang der Sequenz ein.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)

Type Parameters

TSource

Der Typ der Elemente von source.The type of the elements of source.

Parameters

source
IEnumerable<TSource>

Eine Sequenz von Werten.A sequence of values.

element
TSource

Der source voranzustellende Wert.The value to prepend to source.

Returns

IEnumerable<TSource>

Eine neue Sequenz, die mit element beginnt.A new sequence that begins with element.

Exceptions

source ist nullsource is null.

Examples

Im folgenden Codebeispiel wird veranschaulicht, wie mit Prepend ein Wert am Anfang der Sequenz vorangestellt wird.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

Remarks

Note

Diese Methode ändert nicht die Elemente der Auflistung.This method does not modify the elements of the collection. Stattdessen wird eine Kopie der Auflistung mit dem neuen Element erstellt.Instead, it creates a copy of the collection with the new element.

Applies to