Summary: in this tutorial, you will learn how to use the LINQ SkipLast()
method to bypass a specified number of elements from the end of a sequence.
Introduction to the LINQ SkipLast() method
The LINQ SkipLast()
method allows you to skip the last count
elements from a sequence and return the remaining elements after skipping.
Here’s the syntax for using the SkipLast()
method:
IEnumerable<T> SkipLast<TSource>(
this IEnumerable<TSource> source,
int count
)
Code language: C# (cs)
In this syntax:
source
is the input sequence of elements.count
is the number of elements to skip.
The SkipLast()
method returns an IEnumerable<TSource>
of the elements after skipping the last count
elements.
LINQ SkipLast() method example
The following example uses the LINQ SkipLast()
to skip the last three numbers in a list of integers:
using static System.Console;
var numbers = new List<int> { 1, 2, 3, 4, 5, 6, 7 };
var result = numbers.SkipLast(3);
foreach (var number in result)
{
WriteLine(number);
}
Code language: C# (cs)
Output:
1
2
3
4
Code language: C# (cs)
Summary
- Use the LINQ
SkipLast()
method to bypass a specified number of elements from the end of a sequence and return the remaining elements after skipping.
Was this tutorial helpful ?