使用具有泛型接口的泛型实现

Question

I have an interface Value and a class Record

public interface Value<T> {
    T getRawValue();
}

public class Record<Value<T>> {

    private Value<T> value;     

    T getRecordRawValue() {
        return value.getRawValue();
    }

}

Java will not compile this and complains about the Generics declaration > . Could someone enlighten me as to why this is invalid syntax?

Answer 1

You need bound your generic type to indicate the requirement of it being a Value<T> , and you also need to preserve the type that value Value<T> is providing, therefore your Record class requires two generic types: T: The type of the values returned and U: the type that represents a Value<T>

public class Record<T, U extends Value<T>>

Here you have a full example of this:

public interface Value<T> {
    T getRawValue();
}

public class Record<T, U extends Value<T>> {

    public Record(U value) {
        this.value = value;
    }

    private U value;

    T getRecordRawValue() {
        return value.getRawValue();
    }

}

public class StringValue implements Value<String> {

    @Override
    public String getRawValue() {
        return "raw";
    }
}

public class StrongValue implements Value<String> {

    @Override
    public String getRawValue() {
        return "Rrrrraaawww";
    }
}

public class StringRecord extends Record<String, StringValue> {

    public StringRecord(StringValue valueProvider) {
        super(valueProvider);
    }

    public String report() {
        return super.getRecordRawValue();
    }
}

Answer 2

The code public class Record<Value<T>> attempts to declare a generic type parameter for the class called Value<T> , but when declaring a generic type parameter, it should be a simple identifier such as T .

Try

public class Record<T>

Answer 3

Your Record class is wrongly declared. It should be

public class Record<T> {
    //contents...
}