Enumerable.Prepend<TSource>(IEnumerable<TSource>, TSource) Method
Definition
Important
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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
The type of the elements of source
.
Parameters
- source
- IEnumerable<TSource>
A sequence of values.
- element
- TSource
The value to prepend to source
.
Returns
A new sequence that begins with element
.
Exceptions
source
is null
.
Examples
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
This method does not modify the elements of the collection. Instead, it creates a copy of the collection with the new element.
Applies to
.NET
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