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?

Answer 1

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>()

Related Link

Answer 2

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);
}

Answer 3

Change with below code it compile and run perfectly.

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

Answer 4

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