ConcurrentSkipListSet floor() method in Java
Last Updated :
21 Sep, 2018
The floor() method of java.util.concurrent.ConcurrentSkipListSet is an in-built function in Java which returns the greatest element in this set less than or equal to the given element, or null if there is no such element.
Syntax:
ConcurrentSkipListSet.floor(E e)
Parameter: The function accepts a single parameter e i.e. the value to match.
Return Value: The function returns the greatest element less than or equal to e, or null if there is no such element.
Exception: The function throws the following exceptions:
ClassCastException – if the specified element cannot be compared with the elements currently in the set.
NullPointerException – if the specified element is null
Below programs illustrate the ConcurrentSkipListSet.floor() method:
Program 1:
import java.util.concurrent.*;
class ConcurrentSkipListSetFloorExample1 {
public static void main(String[] args)
{
ConcurrentSkipListSet<Integer>
Lset = new ConcurrentSkipListSet<Integer>();
for ( int i = 10 ; i <= 50 ; i += 10 )
Lset.add(i);
System.out.println( "The floor of 20 in the set "
+ Lset.floor( 20 ));
System.out.println( "The floor of 39 in the set "
+ Lset.floor( 39 ));
System.out.println( "The floor of 10 in the set "
+ Lset.floor( 9 ));
}
}
|
Output:
The floor of 20 in the set 20
The floor of 39 in the set 30
The floor of 10 in the set null
Program 2: Program to show NullPointerException in floor().
import java.util.concurrent.*;
class ConcurrentSkipListSetFloorExample2 {
public static void main(String[] args)
{
ConcurrentSkipListSet<Integer>
Lset = new ConcurrentSkipListSet<Integer>();
for ( int i = 10 ; i <= 50 ; i += 10 )
Lset.add(i);
try {
System.out.println( "The floor of null in the set "
+ Lset.floor( null ));
}
catch (Exception e) {
System.out.println( "Exception: " + e);
}
}
}
|
Output:
Exception: java.lang.NullPointerException
Reference: https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/ConcurrentSkipListSet.html#floor-E-
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