How to know if List.remove() is "Unsupported"?

Question

I have this:

import java.util.Arrays;
import java.util.List;
import java.util.ArrayList;
public class ListTest {

    public static void main(String[] args) {


        String[] values = { "yes", "no"};
        List<String> aa = Arrays.asList(values);
        System.out.println(aa.getClass().getName());
        aa.remove(0);
    }

}

It gives:

$ java ListTest 
java.util.Arrays$ArrayList
Exception in thread "main" java.lang.UnsupportedOperationException
    at java.util.AbstractList.remove(AbstractList.java:161)
    at ListTest.main(ListTest.java:12)

Question: I understand why I am getting this exception. It is because the ArrayList class from inside Arrays.java is being used which does not have a remove() method. My question is how can someone(any user, like me) know before using that the List they received that it does not contain a remove method?

score 8 · Accepted Answer · answered Sep 26 '14 at 05:50

There is no way to know. All of the List<T> methods that change the list are listed as optional. A subclass can implement them or not. Unfortunately the API does not include a method like isReadOnly(), so there's no way to check if these methods will throw exceptions without calling them.

It is the responsibility of the owner of the read-only list not to pass it to methods that will try to change it.

score 1 · Answer 2 · answered Sep 26 '14 at 05:53

1

As long as the Interface is properly implemented, writing a remove method that throws an Exception is completely legal. So, no you won't know until it breaks... That is when Javadoc comes handy.

answered Sep 26 '14 at 05:53

Desorder

1,489
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16

score 1 · Answer 3 · edited May 23 '17 at 12:21

1

Here is the solution

import java.util.Arrays;
import java.util.List;
import java.util.ArrayList;
class ListTest {
    public static void main(String[] args) {
        String[] values = { "yes", "no"};
        List<String> aa = new ArrayList<>(Arrays.asList(values));
        System.out.println(aa.getClass().getName());
        aa.remove(0);
  }
}

and here is a very good explanation for your problem.

Hope this helps.

edited May 23 '17 at 12:21

Community

1
1

answered Sep 26 '14 at 06:01

codebot

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Did you forget "String" inside <> here : `List aa = new ArrayList<>(Arrays.asList(values));` – Ankur Agarwal Sep 26 '14 at 06:06
1

No no. That is `diamond inferance`. Allows only in Java 7 and higher. You can use it as both ways if you java version is higher than 7. Check this out. http://stackoverflow.com/questions/4166966/what-is-the-point-of-the-diamond-operator-in-java-7 – codebot Sep 26 '14 at 06:08

Ruchira Gayan Ranaweera · Answer 4 · 2014-09-26T05:56:56.847

0

Arrays.asList returns a List wrapper for an array. This wrapper has a fixed size and is directly backed by the array hence there can't be modify methods. So better to keep in mind that else i am not sure how to know it because at least compile time you will not get an error .

If you want to get a Collection from a array to modify you can use

Collection c = new ArrayList(Arrays.asList(values));

edited Sep 26 '14 at 05:56

answered Sep 26 '14 at 05:51

Ruchira Gayan Ranaweera

32,406
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The OP knows this. The question was whether or not there was any way to detect this lack of implementation without knowing the concrete type of the list. – ApproachingDarknessFish Sep 26 '14 at 05:53

score 0 · Answer 5 · answered Sep 29 '14 at 05:29

Well what u have done that is directly return like this. ex. array in easy way something like this String array[] means its returning array now tell me is there anyway to remove that element from that i mean from the array? that is belongs from list only

one possible way is .

import java.util.List;
import java.util.ArrayList;
import java.util.Arrays;

class ListTest {
    public static void main(String[] args) {
        String[] values = { "yes", "no"};
        List<String> aa = new ArrayList<>(Arrays.asList(values)); // this will return arraylist (according that u wanted).
        System.out.println(aa.getClass().getName());
        aa.remove(0); // removes element as of list interface have remove method.
  }
}

How to know if List.remove() is "Unsupported"?

5 Answers5