design-patterns - 分离策略模式返回的类型
问题描述
我遇到了一些代码问题。为了尝试直截了当,我做了一个尽可能简单的类比——请原谅任何语法错误,因为我在编程方面生疏了,我很快就把它放在一起。
这个想法是我有一些实现策略的类。我需要管理它们的对象以某种方式识别操作返回的类型以便处理它(可能没有向下转换) - 不要问,这不在我的控制之下。
我考虑过使用协变返回类型来避免必须显式引用具体实现(即使那时我必须编写一个处理每种类型的函数,但对我来说这似乎没问题,因为所有东西都放在一起并且对于一个新类型你只是添加新功能)。
public abstract class ArtPiece
{
}
public class Painting extends ArtPiece
{
public Painting()
{
}
// paintings stuff here
}
public class Song extends ArtPiece
{
public Song()
{
}
// songs stuff here
}
public class Book extends ArtPiece
{
public Book()
{
}
// books stuff here
}
///////////////////////////////////////////////////
public abstract class Artist
{
public abstract ArtPiece create();
}
public class Painter extends Artist
{
public Painter()
{
}
public Painting create()
{
return new Painting();
}
}
public class Musician extends Artist
{
public Musician()
{
}
public Song create()
{
return new Song();
}
}
public class Writer extends Artist
{
public Writer()
{
}
public Book create()
{
return new Book();
}
}
//////////////////////////////////////////////////
public class ArtCollector
{
private Artist[] artists;
private Painting[] paintings;
private Song[] songs;
private Book[] books;
private int followedArtists;
private int ownedPaintings;
private int ownedSongs;
private int ownedBooks;
public ArtCollector()
{
followedArtists = 0;
ownedPaintings = 0;
ownedSongs = 0;
ownedBooks = 0;
artists = new Artist[3];
paintings = new Painting[3];
songs = new Song[3];
books = new Book[3];
}
public void followArtist(Artist a)
{
if (followedArtists < artists.length)
{
artists[followedArtists++] = a;
return;
}
System.out.println("Too many artists to follow!");
}
public void commissionArt(int id)
{
storeArtPiece(artists[id].create());
}
private void storeArtPiece(Painting p)
{
if (ownedPaintings < paintings.length)
{
paintings[ownedPaintings++] = p;
return;
}
System.out.println("Too many paintings to store!");
}
private void storeArtPiece(Song s)
{
if (ownedSongs < songs.length)
{
songs[ownedSongs++] = s;
return;
}
System.out.println("Too many songs to store!");
}
private void storeArtPiece(Book b)
{
if (ownedBooks < books.length)
{
books[ownedBooks++] = b;
return;
}
System.out.println("Too many books to store!");
}
public void countPaintings()
{
System.out.println("I own " + paintings.length.toString() + " paintings.");
}
public void countSongs()
{
System.out.println("I own " + songs.length.toString() + " songs.");
}
public void countBooks()
{
System.out.println("I own " + books.length.toString() + " books.");
}
}
//////////////////////////////////////////////////
public static void main(string[] args)
{
ArtCollector collector = new ArtCollector();
collector.followArtist(new Musician());
collector.followArtist(new Painter());
collector.followArtist(new Writer());
collector.commissionArt(0);
collector.commissionArt(0);
collector.commissionArt(1);
collector.commissionArt(2);
collector.commissionArt(1);
collector.commissionArt(0);
collector.commissionArt(0);
collector.countPaintings();
collector.countSongs();
collector.countBooks();
}
首先,我没有测试过这种方法。这只是我必须处理这个问题的一个想法。
我想知道你对此的看法——我在正确的轨道上吗?
你认为有更好的方法来处理这个问题吗?
解决方案
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