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

Tanım

Dizinin başlangıcına bir değer ekler.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)

Tür Parametreleri

TSource

Öğelerinin türü source .The type of the elements of source.

Parametreler

source
IEnumerable<TSource>

Bir dizi değer.A sequence of values.

element
TSource

Eklenecek değer bitiş değeri source .The value to prepend to source.

Döndürülenler

IEnumerable<TSource>

İle başlayan yeni bir sıra element .A new sequence that begins with element.

Özel durumlar

source, null değeridir.source is null.

Örnekler

Aşağıdaki kod örneği, Prepend dizinin başlangıcına bir değer eklemek için nasıl kullanılacağını gösterir.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

Açıklamalar

Not

Bu yöntem, koleksiyonun öğelerini değiştirmez.This method does not modify the elements of the collection. Bunun yerine, yeni öğesiyle koleksiyonun bir kopyasını oluşturur.Instead, it creates a copy of the collection with the new element.

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