java generics - Comparable 类型中的方法compareTo(capture#1-of ?) 不适用于参数

Question

所以我想这样做:

public interface IFieldObject {
    public Comparable get();
}

public interface IFieldCondition {
    public boolean apply(IFieldObject field, Comparable compare);
}

public class EqualTo implements IFieldCondition {
    public boolean apply(IFieldObject field, Comparable compare) {
        return (field.get().compareTo(compare) == 0);       
    }    
}

但是 Eclipse 给了我警告:

Type safety: The method compareTo(Object) belongs to the raw type Comparable. References to generic type Comparable should be parameterized

所以我把它变成了:

public interface IFieldObject {
    public Comparable<?> get();
}

public interface IFieldCondition {
    public boolean apply(IFieldObject field, Comparable<?> compare);
}

public class EqualTo implements IFieldCondition {
    public boolean apply(IFieldObject field, Comparable<?> compare) {
        return (field.get().compareTo(compare) == 0);       
    }
}

由于以下原因无法编译:

The method compareTo(capture#1-of ?) in the type Comparable is not applicable for the arguments (Comparable)

正确的做法是什么？(遵循惯用的 Java >= 1.6，没有警告)

Answer 1

目前您无法保证 field.get() 返回的类型确实与该方法指定的类型具有可比性。理想情况下，使整个事情通用，例如:

public interface IFieldObject<T extends Comparable<T>> {
    public T get();
}

public interface IFieldCondition<T> {
    public boolean apply(IFieldObject<T> field, Comparable<T> compare);
}

public class EqualTo<T> implements IFieldCondition<T> {
    public boolean apply(IFieldObject<T> field, Comparable<T> compare) {
        return (field.get().compareTo(compare) == 0);       
    }
}

毫无疑问，您可以使用额外的捕获来使其更加通用，但这就是起点。