C# 3.0下有限狀態(tài)機的一種優(yōu)雅的實現

實現狀態(tài)機有多種模式，其中最靈活而強大的方式是通過遷移表來實現，該方式的缺點之一是需要編寫大量小塊代碼去支持遷移表。而在C#3.0中，可以以一種非常優(yōu)雅的方式實現。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace StateMachine
{
????class Program
????{
????????static void Main(string[] args)
????????{
????????????var door = new Door(State.Open);

????????????while (true)
????????????{
????????????????string s = Console.ReadLine();
????????????????Operation op = string.IsNullOrEmpty(s) ? Operation.Push : Operation.Pull;
????????????????door.Process(op);
????????????}
????????}
????}

????enum Operation
????{
????????Push, Pull
????}

????enum State
????{
????????Open, Closed
????}

????class Door
????{
????????public State State { get; set; }

????????Dictionary<State, Dictionary<Operation, Action>> rule;
????????public Door(State state)
????????{
????????????this.State = state;

????????????rule = new Dictionary<State, Dictionary<Operation, Action>>();
????????????foreach (var e in Enum.GetValues(typeof(State)))
????????????{
????????????????rule[(State)e] = new Dictionary<Operation, Action>();
????????????}

????????????InitOperationRule();
????????}

????????void InitOperationRule()
????????{
????????????//正常操作
????????????rule[State.Closed][Operation.Push] = () => { Console.WriteLine("門被推開了"); State = State.Open; };
????????????rule[State.Open][Operation.Pull] = () => { Console.WriteLine("門被拉上了"); State = State.Closed; };

????????????加入幾種特殊情況的處理
????????????//rule[State.Closed][Operation.Pull] = () => Console.WriteLine("門是關上的，拉了也白拉");
????????????//rule[State.Open][Operation.Push] = () => Console.WriteLine("門是開的，不用推了，直接進去吧");
????????}

????????public void Process(Operation op)
????????{
????????????try
????????????{
????????????????rule[State][op]();
????????????}
????????????catch (KeyNotFoundException)
????????????{

????????????????Console.WriteLine(string.Format("門在{0}狀態(tài)下不允許{1}操作", State, op));
????????????}
????????????
????????}
????}
}

從代碼中可以看到，通過lambda表達式，可以簡化遷移表的構造，并且更加直觀。

通過遷移表構造狀態(tài)機的一種不足在于查詢速度，在本例中每個操作都要進行兩次查詢才能進行狀態(tài)轉換操作。如果狀態(tài)較多則非常費時，這里我把它改進了一下，使得每次操作只需要查詢一次即可。

????class DoorPlus
????{
????????State state;
????????public State State
????????{
????????????get { return state; }
????????????set
????????????{
????????????????if (state != value)
????????????????????currentOpRule = rule[value];
????????????????state = value;
????????????}
????????}

????????Dictionary<Operation, Action> currentOpRule;
????????Dictionary<State, Dictionary<Operation, Action>> rule;
????????public DoorPlus(State state)
????????{
????????????this.State = state;

????????????rule = new Dictionary<State, Dictionary<Operation, Action>>();
????????????foreach (var e in Enum.GetValues(typeof(State)))
????????????{
????????????????rule[(State)e] = new Dictionary<Operation, Action>();
????????????}

????????????currentOpRule = rule[State];

????????????InitOperationRule();
????????}

????????void InitOperationRule()
????????{
????????????//正常操作
????????????rule[State.Closed][Operation.Push] = () => { Console.WriteLine("門被推開了"); State = State.Open; };
????????????rule[State.Open][Operation.Pull] = () => { Console.WriteLine("門被拉上了"); State = State.Closed; };

????????????加入幾種特殊情況的處理
????????????//rule[State.Closed][Operation.Pull] = () => Console.WriteLine("門是關上的，拉了也白拉");
????????????//rule[State.Open][Operation.Push] = () => Console.WriteLine("門是開的，不用推了，直接進去吧");
????????}

????????public void Process(Operation op)
????????{
????????????try
????????????{
????????????????currentOpRule[op]();
????????????}
????????????catch (KeyNotFoundException)
????????????{

????????????????Console.WriteLine(string.Format("門在{0}狀態(tài)下不允許{1}操作", State, op));
????????????}
????????}
????}

總結

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