Ios DiffableDataSource是否能够正确处理项目移动、项目修改和项目移动&；修改？

我一直觉得，一个好的Diff库，应该具备检测以下3个方面的能力

当项目被移动时

当项目被修改时

移动和修改项目时

---

当我查看WWDC 2019的源代码时，我注意到这些实现只能够检测移动，而不能检测修改

class WiFiController {

    struct Network: Hashable {
        let name: String
        let identifier = UUID()

        func hash(into hasher: inout Hasher) {
            hasher.combine(identifier)
        }
        static func == (lhs: Network, rhs: Network) -> Bool {
            return lhs.identifier == rhs.identifier
        }
    }

移动好啊 检测到移动，因为我们检测到标识符=0从索引0移动到索引1

修改。失败 未检测到修改，因为前后的标识符保持不变

func==

仅考虑标识符

---

我想，我可以通过修改上面的类来解决这个问题

class WiFiController {

    struct Network: Hashable {
        let name: String
        let identifier = UUID()

        func hash(into hasher: inout Hasher) {
            hasher.combine(name, identifier)
        }
        static func == (lhs: Network, rhs: Network) -> Bool {
            return lhs.name == rhs.name && lhs.identifier == rhs.identifier
        }
    }

移动好啊 检测到移动，因为我们检测到name=“WiFi A”标识符=0从索引0移动到索引1

修改。好的（也许） 检测到修改，因为我们检测到name=“WiFi A”标识符=0被修改为name=“WiFi X”标识符=0

或者，图书馆可能会得出结论，我已从索引0中删除name=“WiFi A”标识符=0，并在索引0中插入新的name=“WiFi X”标识符=0

移动和修改。失败 无法检测移动，因为我们正在比较名称和标识符

---

我可以知道你如何解决上述问题吗

我认为这可能是

DiffableDataSource

库的一个缺点。由于仅基于

散列

和

=

，它无法回答这两个问题

2个旧项目和新项目是否相同

如果这两个项目相同，它们的内容是否也相同

当我们实现

func==

时，我们只能使其比较标识或比较内容，但不能两者兼而有之

---

p/s

如果我检查一下安卓系统做得如何，它们提供了两种方法

比较识别

比较内容

/**
 * Called by the DiffUtil to decide whether two object represent the same Item.
 * <p>
 * For example, if your items have unique ids, this method should check their id equality.
 *
 * @param oldItemPosition The position of the item in the old list
 * @param newItemPosition The position of the item in the new list
 * @return True if the two items represent the same object or false if they are different.
 */
public abstract boolean areItemsTheSame(int oldItemPosition, int newItemPosition);

/**
 * Called by the DiffUtil when it wants to check whether two items have the same data.
 * DiffUtil uses this information to detect if the contents of an item has changed.
 * <p>
 * DiffUtil uses this method to check equality instead of {@link Object#equals(Object)}
 * so that you can change its behavior depending on your UI.
 * For example, if you are using DiffUtil with a
 * {@link RecyclerView.Adapter RecyclerView.Adapter}, you should
 * return whether the items' visual representations are the same.
 * <p>
 * This method is called only if {@link #areItemsTheSame(int, int)} returns
 * {@code true} for these items.
 *
 * @param oldItemPosition The position of the item in the old list
 * @param newItemPosition The position of the item in the new list which replaces the
 *                        oldItem
 * @return True if the contents of the items are the same or false if they are different.
 */
public abstract boolean areContentsTheSame(int oldItemPosition, int newItemPosition);

/**
*由DiffUtil调用以确定两个对象是否表示同一项。
*
*例如，如果项目具有唯一的id，则此方法应检查其id是否相等。
*
*@param oldItemPosition项目在旧列表中的位置
*@param newItemPosition项目在新列表中的位置
*@如果两个项目代表相同的对象，则返回True；如果两个项目不同，则返回false。
*/
公共抽象布尔值是相同的（int-oldItemPosition，int-newItemPosition）；
/**
*当DiffUtil想要检查两个项是否具有相同的数据时，调用该函数。
*DiffUtil使用此信息检测项目的内容是否已更改。
*
*DiffUtil使用此方法检查相等性，而不是{@link Object#equals（Object）}
*因此，您可以根据您的UI更改其行为。
*例如，如果将DiffUtil与
*{@link RecyclerView.Adapter RecyclerView.Adapter}，您应该
*返回项目的视觉表示是否相同。
*
*仅当{@link#areItemsTheSame（int，int）}返回时才调用此方法
*{@code true}用于这些项目。
*
*@param oldItemPosition项目在旧列表中的位置
*@param newItemPosition项目在新列表中的位置，该列表将替换
*旧项目
*@如果项目内容相同，则返回True；如果项目内容不同，则返回false。
*/
公共抽象布尔值是相同的内容（int oldItemPosition，int newItemPosition）；

class WiFiController {

    struct Network: Hashable {
        let name: String
        let identifier = UUID()

        func hash(into hasher: inout Hasher) {
            hasher.combine(name, identifier)
        }
        static func == (lhs: Network, rhs: Network) -> Bool {
            return lhs.name == rhs.name && lhs.identifier == rhs.identifier
        }
    }

Index   Before                                      After               
0       [name = "WiFi A", identifier = 0]           [name = "WiFi B", identifier = 1]    
1       [name = "WiFi B", identifier = 1]           [name = "WiFi A", identifier = 0]
2       [name = "WiFi C", identifier = 2]           [name = "WiFi C", identifier = 2]

Index   Before                                      After               
0       [name = "WiFi A", identifier = 0]           [name = "WiFi X", identifier = 0]    
1       [name = "WiFi B", identifier = 1]           [name = "WiFi B", identifier = 1]
2       [name = "WiFi C", identifier = 2]           [name = "WiFi C", identifier = 2]

Index   Before                                      After               
0       [name = "WiFi A", identifier = 0]           [name = "WiFi B", identifier = 1]    
1       [name = "WiFi B", identifier = 1]           [name = "WiFi X", identifier = 0]
2       [name = "WiFi C", identifier = 2]           [name = "WiFi C", identifier = 2]

/**
 * Called by the DiffUtil to decide whether two object represent the same Item.
 * <p>
 * For example, if your items have unique ids, this method should check their id equality.
 *
 * @param oldItemPosition The position of the item in the old list
 * @param newItemPosition The position of the item in the new list
 * @return True if the two items represent the same object or false if they are different.
 */
public abstract boolean areItemsTheSame(int oldItemPosition, int newItemPosition);

/**
 * Called by the DiffUtil when it wants to check whether two items have the same data.
 * DiffUtil uses this information to detect if the contents of an item has changed.
 * <p>
 * DiffUtil uses this method to check equality instead of {@link Object#equals(Object)}
 * so that you can change its behavior depending on your UI.
 * For example, if you are using DiffUtil with a
 * {@link RecyclerView.Adapter RecyclerView.Adapter}, you should
 * return whether the items' visual representations are the same.
 * <p>
 * This method is called only if {@link #areItemsTheSame(int, int)} returns
 * {@code true} for these items.
 *
 * @param oldItemPosition The position of the item in the old list
 * @param newItemPosition The position of the item in the new list which replaces the
 *                        oldItem
 * @return True if the contents of the items are the same or false if they are different.
 */
public abstract boolean areContentsTheSame(int oldItemPosition, int newItemPosition);