不可变类很棒，但有一个大问题我想不出一种合理的方法来解决 - 循环。

class Friend {
   Set<Friend> friends();
}

一个人如何模拟我有你作为 friend ，而我又有我作为 friend ？

不变性 这个来自外界的类肯定是​​不可变的。为了进行相等性检查，内部保存的值应该保持不变。

最佳答案

[[[ 编辑:添加代码以演示完全不可变的概念]]]

这就是构建器非常适合不可变对象(immutable对象)的原因 - 它们允许在构建过程中使用可变性，以便在您“卡住”它之前设置好所有内容。在这种情况下，我猜你需要一个支持创建循环的 friend 生成器。

final FriendBuilder john = new FriendBuilder().setName("john");
final FriendBuilder mary = new FriendBuilder().setName("mary");
final FriendBuilder susan = new FriendBuilder().setName("susan");
john
  .likes(mary)
  .likes(susan);
mary
   .likes(susan)
   .likes(john);
susan
   .likes(john);

// okay lets build the immutable Friends
Map<Friend> friends = FriendsBuilder.createCircleOfFriends(john, mary, susan);
Friend immutableJohn = friends.get("john");

编辑:在下面添加了不可变示例以演示方法:

• 评论中有一些关于不可变版本是否可行的讨论。

• 字段是最终的和不可变的。构造函数中使用了可修改的集合，但构造后只保留不可修改的引用。

• 我有另一个版本，它使用 Guava ImmutableSet 作为真正不可变的集合，而不是 JDK 的不可修改的包装器。它的工作原理相同，但使用了 Guava 的漂​​亮集合构建器。

代码:

import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.IdentityHashMap;
import java.util.Map;
import java.util.Set;

/**
 * Note: potentially cycle graph - be careful of deep equals/hashCode/toString/etc.
 * Immutable
 */
public class Friend {

    public static class Builder {

        private final String name;
        private final Set<Builder> friends =
            new HashSet<Builder>();

        Builder(final String name) {
            this.name = name;
        }

        public String getName() {
            return name;
        }

        public Set<Builder> getFriends() {
            return friends;
        }

        void likes(final Builder... newFriends) {
            for (final Builder newFriend : newFriends)
            friends.add(newFriend);
        }

        public Map<String, Friend> createCircleOfFriends() {
            final IdentityHashMap<Builder, Friend> existing =
                new IdentityHashMap<Builder, Friend>();

            // Creating one friend creates the graph
            new Friend(this, existing);
            // after the call existingNodes contains all the nodes in the graph

            // Create map of the all nodes
            final Map<String, Friend> map =
                new HashMap<String, Friend>(existing.size(), 1f);
            for (final Friend current : existing.values()) {
                map.put(current.getName(), current);
            }

            return map;
        }
    }

    final String name;
    final Set<Friend> friends;

    private Friend(
            final Builder builder,
            final Map<Builder, Friend> existingNodes) {
        this.name = builder.getName();

        existingNodes.put(builder, this);

        final IdentityHashMap<Friend, Friend> friends =
            new IdentityHashMap<Friend, Friend>();
        for (final Builder current : builder.getFriends()) {
            Friend immutableCurrent = existingNodes.get(current);
            if (immutableCurrent == null) {
                immutableCurrent =
                    new Friend(current, existingNodes);
            }
            friends.put(immutableCurrent, immutableCurrent);
        }

        this.friends = Collections.unmodifiableSet(friends.keySet());
    }

    public String getName() {
        return name;
    }

    public Set<Friend> getFriends() {
        return friends;
    }


    /** Create string - prints links, but does not traverse them */
    @Override
    public String toString() {
        final StringBuffer sb = new StringBuffer();
        sb.append("Friend ").append(System.identityHashCode(this)).append(" {\n");
        sb.append("  name = ").append(getName()).append("\n");
        sb.append("  links = {").append("\n");
        for (final Friend friend : getFriends()) {
            sb
            .append("     ")
            .append(friend.getName())
            .append(" (")
            .append(System.identityHashCode(friend))
            .append(")\n");
        }
        sb.append("  }\n");
        sb.append("}");
        return sb.toString();
    }

    public static void main(final String[] args) {
        final Friend.Builder john = new Friend.Builder("john");
        final Friend.Builder mary = new Friend.Builder("mary");
        final Friend.Builder susan = new Friend.Builder("susan");
        john
          .likes(mary, susan);
        mary
           .likes(susan, john);
        susan
           .likes(john);

        // okay lets build the immutable Friends
        final Map<String, Friend> friends = john.createCircleOfFriends();

        for(final Friend friend : friends.values()) {
            System.out.println(friend);
        }

        final Friend immutableJohn = friends.get("john");
    }
}

输出:

Node 11423854 {
  value = john
  links = {
     susan (19537476)
     mary (2704014)
  }
}
Node 2704014 {
  value = mary
  links = {
     susan (19537476)
     john (11423854)
  }
}
Node 19537476 {
  value = susan
  links = {
     john (11423854)
  }
}

关于java - 如何对不可变类实例之间的循环建模？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/4834513/