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Home/ Questions/Q 834453
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T04:40:24+00:00 2026-05-15T04:40:24+00:00

I have classes as below. public interface ITest <T> { public Set<T> methodHere(); }

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I have classes as below.

public interface ITest <T>
{
   public Set<T> methodHere();
}

public class test1 implements ITest<String>
{
   Set<String> methodHere(){
      return // Set of String
   }
}

public class test2 implements ITest<Integer>
{
   Set<Integer> methodHere(){
     return // Set of Integer
   }
}

public class ITestFactory {
 public static ITest getInstance(int type) {
  if(type == 1) return new test1();
  else if(type == 2) return new test2();
 }
}
public class TestUser {
    public doSomething(int type) {
       ITest t = ITestFactory.getInstance(type);
       if(type == 1) Set<Integer> i = t.methodHere();
       else if(type == 2) Set<String> s = t.methodHere();
       ...
    }
}

There is a warning in the factory class that ITest is used as raw type. What modification should I do to get rid of it?

The TestUser code looks ugly. Am I missing something very basic? I don’t want to use Set<?>

Thanks
Nayn

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T04:40:25+00:00Added an answer on May 15, 2026 at 4:40 am

    You can return a ITest<?> to get rid of the warning but probably you want a more strongly type aproach:

    public class TestFactory {
       public static ITest<?> getInstance(int type) {
          if(type == 1) 
             return new test1();
          else if(type == 2) 
             return new test2();
          else
             throw new IllegalArgumentException("Unknown type");
       }
    
       public static <T> ITest<T> getInstance(Class<T> clazz) {
          if(clazz == String.class) 
             return new test1();
          else if(clazz == Integer.class) 
             return new test2();
          else 
             throw new IllegalArgumentException("Unknown type");
       }
    }
    
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