Solved

Remove type safety warning:

Posted on 2006-07-02
7
495 Views
Last Modified: 2008-02-01
The following gives a type safety warning which I am unable to lose, what is wrong?

import java.util.ArrayList;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Random;
public class ArrayL
{
     public static void main(String[] args)
     {
       
          Scanner Input = new Scanner(System.in);
          int Cnt = Input.nextInt();
          ArrayList[] Hand = new ArrayList[Cnt];
          for(int i=0;i<Cnt;i++)
          {
             Hand[i] = new ArrayList<Integer>();
               for(int a =0;a<5;a++)
               {
                    Random Num = new Random();
                    int NumC= Num.nextInt()%4;
                    Hand[i].add(NumC);        ///////////WARNING
               }
          }
          for(int i=0;i<Cnt;i++)
          {
               Iterator T = Hand[i].iterator();
               while (T.hasNext())
               {
                    System.out.print(T.next() + ",");
               }
               System.out.println();
          }
     }
}

Type safety: The method add(Object) belongs to the raw type ArrayList. References to generic type ArrayList<E> should be parameterized
0
Comment
Question by:List244
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
  • 2
7 Comments
 
LVL 23

Expert Comment

by:Ajay-Singh
ID: 17027533
> ArrayList[] Hand = new ArrayList[Cnt];
should be

ArrayList<Integer>[] Hand = new ArrayList<Integer>[Cnt];
0
 
LVL 8

Author Comment

by:List244
ID: 17027539
Tried that, gives:

Cannot create a generic array of ArrayList<Integer>
0
 
LVL 23

Expert Comment

by:Ajay-Singh
ID: 17027644
May not be able to do this:
http://www.cis.upenn.edu/~matuszek/cit594-2006/Pages/coping-with-generics.html

You can try this:
                ArrayList<ArrayList<Integer>> l = new ArrayList<ArrayList<Integer>>(Cnt);
                ArrayList<Integer>[] Hand = (ArrayList<Integer>[]) l.toArray();

Why don't you use ArrayList<ArrayList<Integer>> instead of ArrayList<Integer>[]
0
SharePoint Admin?

Enable Your Employees To Focus On The Core With Intuitive Onscreen Guidance That is With You At The Moment of Need.

 
LVL 8

Author Comment

by:List244
ID: 17027697
This adds the following warning:

Type safety: The cast from Object[] to ArrayList<Integer>[] is actually checking against the erased type ArrayList[]
0
 
LVL 92

Accepted Solution

by:
objects earned 500 total points
ID: 17028045
Don't think you can get rid of that waring using an array (as you cannot create a array of generics), would instead need to use collection

import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Random;
import java.util.Scanner;
public class ArrayL
{
     public static void main(String[] args)
     {
       
          Scanner Input = new Scanner(System.in);
          int Cnt = Input.nextInt();
          List<List<Integer>> Hand = new ArrayList<List<Integer>>(Cnt);
          for(int i=0;i<Cnt;i++)
          {
             Hand.add(new ArrayList<Integer>());
               for(int a =0;a<5;a++)
               {
                    Random Num = new Random();
                    int NumC= Num.nextInt()%4;
                    Hand.get(i).add(NumC);        ///////////WARNING
               }
          }
          for(int i=0;i<Cnt;i++)
          {
               Iterator T = Hand.get(i).iterator();
               while (T.hasNext())
               {
                    System.out.print(T.next() + ",");
               }
               System.out.println();
          }
     }
}
0
 
LVL 92

Expert Comment

by:objects
ID: 17028060
your other option if you want to use array it to suppress the warning:

import java.util.ArrayList;
import java.util.Iterator;
import java.util.Scanner;
import java.util.Random;
public class ArrayL
{
      @SuppressWarnings("unchecked")
     public static void main(String[] args)
     {
       
          Scanner Input = new Scanner(System.in);
          int Cnt = Input.nextInt();
          ArrayList[] Hand = new ArrayList[Cnt];
          for(int i=0;i<Cnt;i++)
          {
             Hand[i] = new ArrayList<Integer>();
               for(int a =0;a<5;a++)
               {
                    Random Num = new Random();
                    int NumC= Num.nextInt()%4;
                    Hand[i].add(NumC);
               }
          }
          for(int i=0;i<Cnt;i++)
          {
               Iterator T = Hand[i].iterator();
               while (T.hasNext())
               {
                    System.out.print(T.next() + ",");
               }
               System.out.println();
          }
     }
}
0
 
LVL 8

Author Comment

by:List244
ID: 17028118
Thanks, that worked for me.
0

Featured Post

Get 15 Days FREE Full-Featured Trial

Benefit from a mission critical IT monitoring with Monitis Premium or get it FREE for your entry level monitoring needs.
-Over 200,000 users
-More than 300,000 websites monitored
-Used in 197 countries
-Recommended by 98% of users

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java contains several comparison operators (e.g., <, <=, >, >=, ==, !=) that allow you to compare primitive values. However, these operators cannot be used to compare the contents of objects. Interface Comparable is used to allow objects of a cl…
In this post we will learn different types of Android Layout and some basics of an Android App.
Viewers will learn about basic arrays, how to declare them, and how to use them. Introduction and definition: Declare an array and cover the syntax of declaring them: Initialize every index in the created array: Example/Features of a basic arr…
This tutorial covers a practical example of lazy loading technique and early loading technique in a Singleton Design Pattern.

729 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question