SkipLast Method
public static IEnumerable<TSource> SkipLast<TSource>(this IEnumerable<TSource> source, int count)
Bypasses a specified number of elements at the end of a sequence and then returns the remaining elements.
Returns
An IEnumerable<TSource> that contains the elements that follow the omitted elements at the end of the input sequence. If count is equal to or greater than the count of elements in source, an empty IEnumerable<TSource> is returned.
Remarks
- This method performs a deferred execution. The values of
sourceandcountare not validated until the result of this method is enumerated. - Negative values of
countare treated as zero. - If the source sequence is empty, an empty sequence is returned.
- The
SkipLastmethod works by first counting the total number of elements in the source sequence. Then, it returns all elements except for the lastcountelements. - This method is the opposite of the
TakeLastmethod.
Example
// Sample usage of SkipLast
int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// Skip the last 3 elements
var skipped = numbers.SkipLast(3);
// The 'skipped' sequence will contain { 1, 2, 3, 4, 5, 6, 7 }
foreach (var number in skipped)
{
Console.WriteLine(number);
}
// Example with an empty sequence
int[] emptyNumbers = { };
var skippedFromEmpty = emptyNumbers.SkipLast(2); // Result is an empty sequence
// Example with count greater than sequence length
var skippedTooMany = numbers.SkipLast(15); // Result is an empty sequence
// Example with negative count
var skippedNegative = numbers.SkipLast(-5); // Treated as SkipLast(0), returns all elements
Parameters
| Name | Description |
|---|---|
source |
An IEnumerable<TSource> to return elements from. |
count |
The number of elements to skip from the end of the sequence. |
Type Parameters
| Name | Description |
|---|---|
TSource |
The type of the elements of source. |