进程同步题（二）哲学家进餐

旧城等待， 2022-09-09 01:48 169阅读 0赞

##### 1.使用C++ std::lock实现多锁原子锁定 #####

只有当一个哲学家同时获得自己左右两边的筷子时，才进餐。

class DiningPhilosophers {
        
    private:
    	std::mutex mutexs[5];
    public:
    	DiningPhilosophers() {
        
    	}
    
    	void wantsToEat(int philosopher) {
        
    		int lhs = philosopher;
    		int rhs = (philosopher + 1) % 5;
    
    		std::lock(mutexs[lhs], mutexs[rhs]);
    		std::lock_guard<std::mutex> lock_a(mutexs[lhs], std::adopt_lock);
    		std::lock_guard<std::mutex> lock_b(mutexs[rhs], std::adopt_lock);
    
    		cout << "哲学家" << philosopher << "进餐" << endl; 
    	}
    };
    
    int main()
    {
        
    	DiningPhilosophers d;
    	thread t0(&DiningPhilosophers::wantsToEat, &d, 0);
    	thread t1(&DiningPhilosophers::wantsToEat, &d, 1);
    	thread t2(&DiningPhilosophers::wantsToEat, &d, 2);
    	thread t3(&DiningPhilosophers::wantsToEat, &d, 3);
    	thread t4(&DiningPhilosophers::wantsToEat, &d, 4);
    
    	t0.join();
    	t1.join();
    	t2.join();
    	t3.join();
    	t4.join();
    
    	return 0;
    }

这里用到了 std::lock()对多个锁进行加锁，同时它是可避免死锁的。  
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std::lock(mutexs[lhs], mutexs[rhs]);
    std::lock_guard<std::mutex> lock_a(mutexs[lhs], std::adopt_lock);
    std::lock_guard<std::mutex> lock_b(mutexs[rhs], std::adopt_lock);

指定锁定策略。

<table> 
 <thead> 
  <tr> 
   <th>类型</th> 
   <th>效果</th> 
  </tr> 
 </thead> 
 <tbody> 
  <tr> 
   <td>defer_lock_t</td> 
   <td>不获得互斥的所有权</td> 
  </tr> 
  <tr> 
   <td>try_to_lock_t</td> 
   <td>尝试获得互斥的所有权而不阻塞</td> 
  </tr> 
  <tr> 
   <td>adopt_lock_t</td> 
   <td>假设调用方线程已拥有互斥的所有权</td> 
  </tr> 
 </tbody> 
</table>

![在这里插入图片描述][watermark_type_ZHJvaWRzYW5zZmFsbGJhY2s_shadow_50_text_Q1NETiBA5oiR5Y-rUlQ_size_20_color_FFFFFF_t_70_g_se_x_16 1]

##### 2.奇偶编号加锁顺序不同 #####

奇数位置的哲学家先拿左手边筷子，偶数位置的哲学家先拿右手边的筷子。这样，奇偶位置的哲学家必然冲突，只有等一方放下筷子，另一方才能拿起。

class DiningPhilosophers {
        
    private:
    	std::mutex mutexs[5];
    public:
    	DiningPhilosophers() {
        
    	}
    
    	void wantsToEat(int philosopher) {
        
    		int lhs = philosopher;
    		int rhs = (philosopher + 1) % 5;
    
    		if (philosopher & 1) {
         //奇数
    			std::lock_guard<std::mutex> lock1(mutexs[lhs]);
    			std::lock_guard<std::mutex> lock2(mutexs[rhs]);
    			cout << "哲学家" << philosopher << "进餐" << endl;
    		}
    		else {
        
    			std::lock_guard<std::mutex> lock2(mutexs[rhs]);
    			std::lock_guard<std::mutex> lock1(mutexs[lhs]);
    			cout << "哲学家" << philosopher << "进餐" << endl;
    		}
    	}
    };

3.限定同时就餐的哲学家数量  
同一时刻，限定只有n-1个哲学家进餐，这样必然不会死锁，因为总会有一个哲学家能够吃到食物，等他吃完，其他哲学家就有机会继续进餐。

class DiningPhilosophers {
        
    private:
    	counting_semaphore<4> sem;
    	mutex mutexs[5];
    public:
    	DiningPhilosophers():sem(4) {
        
    	}
    
    	void wantsToEat(int philosopher) {
        
    		int lhs = philosopher;
    		int rhs = (philosopher + 1) % 5;
    
    		sem.acquire();	// 最多同时4人可同时进餐，最坏都拿同一边筷子，那么至少有一人可以进餐
    		{
        
    			std::lock_guard<std::mutex> lock1(mutexs[lhs]);
    			std::lock_guard<std::mutex> lock2(mutexs[rhs]);
    
    			cout << "哲学家" << philosopher << "进餐" << endl;
    		}
    		sem.release();
    	}
    };

4.指定一个人和其他人拿筷子的顺序不一样  
只要有一个哲学家最开始的拿筷子顺序与其他哲学家**都**不同，必然不会产生死锁。

class DiningPhilosophers {
        
    private:
    	std::mutex mutexs[5];
    public:
    	DiningPhilosophers() {
        
    	}
    
    	void wantsToEat(int philosopher) {
        
    		int lhs = philosopher;
    		int rhs = (philosopher + 1) % 5;
    
    		if (philosopher == 1) {
         //指定一个人和其他人方向不一样 五个人都可以
    			std::lock_guard<std::mutex> lock1(mutexs[lhs]);
    			std::lock_guard<std::mutex> lock2(mutexs[rhs]);
    			cout << "哲学家" << philosopher << "进餐" << endl;
    		}
    		else {
        
    			std::lock_guard<std::mutex> lock2(mutexs[rhs]);
    			std::lock_guard<std::mutex> lock1(mutexs[lhs]);
    			cout << "哲学家" << philosopher << "进餐" << endl;
    		}
    	}
    };

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