如何创建正确的返回类型？

Question

我会在底部发布我的问题。

下面是其他类扩展的类。

public class people {

    class family extends people {   
    }

    class friends extends people {  
    }

    class coworkers extends people {    
    }
}

下面是具有 getAllPeopleByClass 方法的类，该方法由选择类中的 getMembers() 方法调用：

public class processing {

static processing process = null;
private Collection<family> familyList = new ArrayList<family>();
private Collection<coworkers> cowList = new ArrayList<coworkers>();
private Collection<friends> friendList = new ArrayList<friends>();

public processing(){
}

public static processing getInstance() {
    if (process == null)
        process = new processing();
    return process;
}

public <T> Collection<people> getAllPeopleByClass(Class<T> clazz) {
    Collection<people> peopleCollection;
    peopleCollection.addAll(getList(clazz));
    return peopleCollection;
}

private <T> Collection<? extends people> getList(Class<T> clazz) {
    if (clazz == family.class) {  
        return familyList;  
    } else if (clazz == coworkers.class) {
        return cowList;
    } else { // else if clazz == friends.class
        return friendList;
    }
}

最后，使用处理的类：

 public class familySelection {
  public Collection<family> getMembers() {
    Collection<family> f;
    f = processing.getInstance().getAllPeopleByClass(family.class); //type mismatch
    return f; 
  }
}

 public class coworkerSelection {
  public Collection<coworkers> getMembers() {
    Collection<coworkers> c;
    c = processing.getInstance().getAllPeopleByClass(coworkers.class); //type mismatch
    return c;
  }
}


public class friendsSelection {
  public Collection<friends> getMembers() {
    Collection<friends> f;
    f = processing.getInstance().getAllPeopleByClass(friends.class); //type mismatch
    return f;
  }
}

我的问题是我在每个 getMembers() 方法中的 getAllPeopleByClass 调用中都出现了类型不匹配。

我试过这个：

public class familySelection {
  public Collection<family> getMembers() {
    Collection<? extends people> f; //changed to extend from people
    f = processing.getInstance().getAllPeopleByClass(family.class);
    return (Collection<family>) f; //cast warning - dont want that
  }
}

这可行，但我收到一个演员警告，我不想也不想压制它。这是我最接近解决问题的方法。此外，getMembers() 的返回类型必须保持不变。有没有人有办法在没有任何警告的情况下做到这一点？或者甚至以某种方式一般地处理这个？谢谢！

Answer 1

public <T> Collection<T> getAllPeopleByClass(Class<T> clazz) {
    Collection<T> peopleCollection = new ArrayList<T>();
    for(people p : getList(clazz)) {
      peopleCollection.add(clazz.cast(p));
    }
    return peopleCollection;
}

Answer 2

使用super 关键字：

public class familySelection {
  public Collection<? super family> getMembers() {
    Collection<? super family> f;
    f = processing.getInstance().getAllPeopleByClass(family.class);
    return f;
  }
}

UPD： 使用super 通配符，当您只将put 值转换为结构时。当您仅在结构中使用 get 值时，请使用extends 通配符。同时使用get 和put 时不要使用通配符。

Answer 3

我会选择这个版本。它还需要警告抑制，但我认为这是更好的地方。

我还允许自己更改您的 Singleton 实现。如果您真的想这样做，请考虑使用枚举。

public enum processing {
    INSTANCE;

    private Collection<family> familyList = new ArrayList<family>();
    private Collection<coworkers> cowList = new ArrayList<coworkers>();
    private Collection<friends> friendList = new ArrayList<friends>();

    public static processing getInstance() {
        return INSTANCE;
    }

    public <T extends people> Collection<T> getAllPeopleByClass(Class<T> clazz) {
        Collection<T> peopleCollection = new ArrayList<T>();
        peopleCollection.addAll(getList(clazz));
        return peopleCollection;
    }

    @SuppressWarnings("unchecked")
    private <T extends people> Collection<T> getList(Class<T> clazz) {
        if (clazz == family.class) {
            return (Collection<T>) familyList;
        } else if (clazz == coworkers.class) {
            return (Collection<T>) cowList;
        } else { // else if clazz == friends.class
            return (Collection<T>) friendList;
        }
    }
}

和familySelection：

public class familySelection {
    public Collection<family> getMembers() {
        Collection<family> f = processing.getInstance().getAllPeopleByClass(family.class);
        return f;
    }
}