MoreLinq
Concat<(Of <(<'T>)>)> Method (head, tail)
NamespacesMoreLinqMoreEnumerableConcat<(Of <<'(T>)>>)(IEnumerable<(Of <<'(T>)>>), T)
Returns a sequence consisting of the head elements and the given tail element.
Declaration Syntax
C#Visual BasicVisual C++
public static IEnumerable<T> Concat<T>(
	this IEnumerable<T> head,
	T tail
)
<ExtensionAttribute> _
Public Shared Function Concat(Of T) ( _
	head As IEnumerable(Of T), _
	tail As T _
) As IEnumerable(Of T)
public:
[ExtensionAttribute]
generic<typename T>
static IEnumerable<T>^ Concat(
	IEnumerable<T>^ head, 
	T tail
)
Generic Template Parameters
T
Type of sequence
Parameters
head (IEnumerable<(Of <(<'T>)>)>)
All elements of the head. Must not be null.
tail (T)
Tail element of the new sequence.
Return Value
A sequence consisting of the head elements and the given tail element.
Usage Note
In Visual Basic and C#, you can call this method as an instance method on any object of type IEnumerable<(Of <(<'T>)>)>. When you use instance method syntax to call this method, omit the first parameter.
Remarks
This operator uses deferred execution and streams its results.

Assembly: MoreLinq (Module: MoreLinq.dll) Version: 1.0.16006.0 (1.0.16006.1845)