如何創建正確的返回類型？

Question

我會在底部發表我的問題。

下面是其他類的擴展類。

public class people { 

    class family extends people { 
    } 

    class friends extends people { 
    } 

    class coworkers extends people {  
    } 
} 

下面是具有得到由getMembers調用（該方法getAllPeopleByClass）方法在選擇的類的類：

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; 
    } 
} 

最後，使用處理的類：

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; 
    } 
} 

我的問題是，我從getAmbers（）方法的每個getAllPeopleByClass調用中得到類型不匹配。

我已經試過這樣：

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 
    } 
} 

這工作，但我得到一個警告投，我不想，我不想壓抑它。這是我最接近解決問題的方法。此外，getMembers（）的返回類型必須保持原樣。有沒有人有辦法做到這一點，沒有任何警告？或者甚至以某種方式處理這一般？謝謝！

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關鍵字：

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

UPD：使用通配符super，當你只put值轉換爲結構。使用extends通配符時，只有get值不在結構中。當您同時使用get和put時，請勿使用通配符。

Answer 3

我會選擇這個版本。它也需要警告抑制，但我認爲這是更好的選擇。

我也允許自己改變你的Singleton實現。如果你真的想這樣做考慮使用枚舉。

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; 
     } 
    } 
} 

而且familySelection：

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