Method Index

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Index<TSource>(IEnumerable<TSource>)

Returns a sequence of tuples where the index is the zero-based index of the item in the source sequence.

Declaration
public static IEnumerable<(int Index, TSource Item)> Index<TSource>(this IEnumerable<TSource> source)
Parameters
Type Name Description
IEnumerable<TSource> source

The source sequence.
Returns
Type Description
IEnumerable<(int Index, TSource Item)>

A sequence of tuples.
Type Parameters
Name Description
TSource

Type of elements in source sequence.
Remarks

This operator uses deferred execution and streams its results.

Examples

The following code example demonstrates how to return an index with each element of a sequence using Index.

var sequence = new[]
{
    (key: "jan", value: 123),
    (key: "Jan", value: 456),
    (key: "JAN", value: 789),
};

// Index a sequence
var result = sequence
    .Index();

Console.WriteLine(
    "[ " +
    string.Join(", ", result) +
    " ]");

// This code produces the following output:
// [ (0, (jan, 123)), (1, (Jan, 456)), (2, (JAN, 789)) ]
Exceptions
Type Condition
ArgumentNullException

source is null.
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Index<TSource>(IEnumerable<TSource>, int)

Returns a sequence of tuples where the index is the startIndex-based index of the item in the source sequence.

Declaration
public static IEnumerable<(int Index, TSource Item)> Index<TSource>(this IEnumerable<TSource> source, int startIndex)
Parameters
Type Name Description
IEnumerable<TSource> source

The source sequence.
int startIndex

The index of the first value returned.
Returns
Type Description
IEnumerable<(int Index, TSource Item)>

A sequence of tuples.
Type Parameters
Name Description
TSource

Type of elements in source sequence.
Remarks

This operator uses deferred execution and streams its results.

Examples

The following code example demonstrates how to return an index with each element of a sequence using Index.

var sequence = new[]
{
    (key: "jan", value: 123),
    (key: "Jan", value: 456),
    (key: "JAN", value: 789),
};

// Index a sequence
var result = sequence
    .Index(5);

Console.WriteLine(
    "[ " +
    string.Join(", ", result) +
    " ]");

// This code produces the following output:
// [ (5, (jan, 123)), (6, (Jan, 456)), (7, (JAN, 789)) ]
Exceptions
Type Condition
ArgumentNullException

source is null.