incorrect

Question

How to fix this, it says

java.lang.UnsupportedOperationException;
null(in java.util.AbstractList) on i.remove();

https://code.sololearn.com/cMfULUxMMU8f

case 5: //Delete by Name
System.out.print("Enter NAME to DELETE: ");
name=console.next();
found = false;

c= Arrays.asList(p);
ListIterator<Person>i = c.listIterator();

while(i.hasNext()){
Person e = i.next();
if (e.getName().equals(name)) {

System.out.println("Record Deleted Successfully: ");
i.remove();
found = true;

}
}

if (!found){
System.out.println("Record Not Found: ");
}

Answer 1

Don't use a fixed size array, use the Collection API throughout.

Answer 2

// Hope this link helps you

Java List remove() Methods – ArrayList remove()

https://www.journaldev.com/31869/java-list-remove-methods-arraylist-remove

Answer 3

yes, It would be better solution to use Arraylist in whole code.
If you don't want to do that,
for remove() you can use own method like

static void arrayRemove( Object[] array, int index) {
/* where copy (shift) all elements
from index+1 to index
index++
then put one null at the end
array[end] = null; */
}

in case 5:
//i.remove();
arrayRemove(p, i.previousIndex() );

( i.remove() doesn't work
because asList(p) internaly uses fixed size p[]
and remove is not supported )
--
another problem in cases 4 and 5 is then
possible null in the array
in the section at the end of array,
so you have to check it before equals()
if ( e != null) {
and break the loop if e is null