instantiating parameterized class

Question

Just wondering if this is the right way to do it. I want to construct instances of my parametrized class where one of the instance variables is a generic type. The code below works but I get a lot of warnings in the main method "SomeObject is a raw type. References to generic type SomeObject should be parameterized".

public class SomeObject<T> {

    private String description;

    private T value;


    public SomeObject(String description, T value) {
        this.description = description;
        this.value = value;
    }



public static void main(String args[]){

    List <SomeObject> objectList = new ArrayList<SomeObject>();

    objectList.add(new SomeObject("Object 1: ", true));
    objectList.add(new SomeObject("Object 2: ", 888.00));
    objectList.add(new SomeObject("Object 3: ", "another object"));
    objectList.add(new SomeObject("Object 4: ", '4'));

    for (SomeObject object : objectList){
    System.out.println(object.getDescription() + object.getValue());
    }
}

}

What part of the error don't you understand? You should actually use your generics. — SLaks
– SLaks, Commented Apr 25, 2017 at 15:27
Try parameterizing the object creation like you have done for arraylist — Amar Dev
– Amar Dev, Commented Apr 25, 2017 at 15:32
"should it be new SomeObject() instead of new Object()? " Yes it should. I edited it. Thanks. — bez
– bez, Commented Apr 25, 2017 at 15:35
Just a side note: in Java 7 and later, you can often omit the type in the second set of angle brackets, and let it be inferred. It is called the diamond operator. — Logan
– Logan, Commented Apr 25, 2017 at 15:37

Ousmane D. · Accepted Answer · 2017-04-25 15:40:42Z

4

The code below works but I get a lot of warnings in the main method "Object is a raw type. References to generic type Object should be parameterized".

the warning is because you haven't specified the type arguments when constructing the SomeObject. ie.

it should be:

List<SomeObject<?>> objectList = new ArrayList<>();
objectList.add(new SomeObject<Boolean>("Object 1: ", true));
objectList.add(new SomeObject<Double>("Object 2: ", 888.00));
objectList.add(new SomeObject<String>("Object 3: ", "another object"));
objectList.add(new SomeObject<Character>("Object 4: ", '4'));

edited Apr 25, 2017 at 15:40

user180100

answered Apr 25, 2017 at 15:38

Ousmane D.

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2 Comments

user180100 Over a year ago

Can you add the declaration of objectList with type to complete your answer?

Ousmane D. Over a year ago

@RC. sure thing.

Community · Accepted Answer · 2017-05-23 11:46:55Z

When you have a SomeObject without a type argument (the part in the square brackets), that is called a raw type, and it's the same as using the erasure of SomeObject. (Basically, the erasure means it's non-generic.)

You also need to provide a type argument to the SomeObject part of the List. Here I've used a wildcard, which means the list can hold any type of SomeObject, but once we put a SomeObject in to the list we don't know what their original type argument was anymore:

List<SomeObject<?>> objectList = new ArrayList<SomeObject<?>>();

objectList.add(new SomeObject<Boolean>("Object 1: ", true));
objectList.add(new SomeObject<Double>("Object 2: ", 888.00));
objectList.add(new SomeObject<String>("Object 3: ", "another object"));
objectList.add(new SomeObject<Character>("Object 4: ", '4'));

for (SomeObject<?> object : objectList) {
    ...;
}

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