Instantiating a Generic Interface

Question

I have an interface

public interface Foo<T> {
    public void bar(String s, T t);
}

I want to write a method

public void baz() {
    String hi = "Hello";
    String bye = "Bye";
    Foo<String> foo = new Foo() {
        public void bar(String s, String t) {
            System.out.println(s);
            System.out.println(s);
        }
    };
    foo.bar(hi,bye);
}

i get an error

<anonymous Test$1> is not abstract and does not override abstract method bar(String,Object) in Foo
    Foo<String> foo = new Foo() {

I'm fairly new to Java, I'm sure this is a simple mistake. how can I write this?

score 6 · Accepted Answer · edited May 23 '17 at 12:27

6

If you are using java 7, Type inference doesn't apply here. You have to provide the Type parameter in the constructor invoking as well.

    Foo<String> foo = new Foo<String>() {
        public void bar(String s, String t) {
            System.out.println(s);
            System.out.println(s);
        }
    };
    foo.bar(hi,bye);

EDIT: just noticed that you have used new Foo() which is basically a raw type, you have to provide the generic type for your constructor invokation, new Foo<String>()

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edited May 23 '17 at 12:27

Community

1
1

answered Jan 24 '13 at 22:07

PermGenError

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1

Except the OP wasn't using the `<>` operator - he was using raw `Foo`. – Paul Bellora Jan 24 '13 at 22:38
Are you saying `new Foo` is the same as `new Foo<>` in Java 7? I wasn't aware of that. – Paul Bellora Jan 24 '13 at 23:09
nopes, i never said that, they arent the same , `new Foo<>`(not for anonynymous class's though) would only work in java 7 due to introduction of type inference. – PermGenError Jan 24 '13 at 23:14
@PaulBellora sorry my explanation was vague, edited it now.. :) – PermGenError Jan 24 '13 at 23:19

score 3 · Answer 2 · answered Jan 24 '13 at 22:08

You have forgotten one <String>

public void baz() {
    String hi = "Hello";
    String bye = "Bye";
    Foo<String> foo = new Foo<String>() {
        public void bar(String s, String t) {
            System.out.println(s);
            System.out.println(s);
        }
    };
    foo.bar(hi,bye);
}

score 0 · Answer 3 · edited Jan 24 '13 at 22:29

0

Foo<String> foo = new Foo<String>() {
  @Override
  public void bar(String s, String t) {
    System.out.println(s);
    System.out.println(s);
  }
};

edited Jan 24 '13 at 22:29

AMIC MING

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answered Jan 24 '13 at 22:07

Mythul

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score 0 · Answer 4 · answered Jan 24 '13 at 22:11

0

Change with below code it compile and run perfectly.

        Foo<String> foo = new Foo<String>() {

answered Jan 24 '13 at 22:11

Ghanshyam Gupta

1

Instantiating a Generic Interface

4 Answers4