MoreLinq
Fold<(Of <(<'T, TResult>)>)> Method (source, folder)
NamespacesMoreLinqMoreEnumerableFold<(Of <<'(T, TResult>)>>)(IEnumerable<(Of <<'(T>)>>), Func<(Of <<'(T, T, TResult>)>>))
Returns the result of applying a function to a sequence of 2 elements.
Declaration Syntax
C#Visual BasicVisual C++
public static TResult Fold<T, TResult>(
	this IEnumerable<T> source,
	Func<T, T, TResult> folder
)
<ExtensionAttribute> _
Public Shared Function Fold(Of T, TResult) ( _
	source As IEnumerable(Of T), _
	folder As Func(Of T, T, TResult) _
) As TResult
public:
[ExtensionAttribute]
generic<typename T, typename TResult>
static TResult Fold(
	IEnumerable<T>^ source, 
	Func<T, T, TResult>^ folder
)
Generic Template Parameters
T
Type of element in the source sequence
TResult
Type of the result
Parameters
source (IEnumerable<(Of <(<'T>)>)>)
The sequence of items to fold.
folder (Func<(Of <(<'T, T, TResult>)>)>)
Function to apply to the elements in the sequence.
Return Value

[Missing <returns> documentation for "M:MoreLinq.MoreEnumerable.Fold``2(System.Collections.Generic.IEnumerable{``0},System.Func{``0,``0,``1})"]

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 immediate execution and effectively buffers as many items of the source sequence as necessary.

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