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

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)

Typparametrar

Parametrar

Returer

En ny sekvens som börjar med element.

Undantag

Exempel

Följande kodexempel visar hur du använder Prepend för att förbereda ett värde till början av sekvensen.

// 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

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