linux下c++的thread问题

clyman

1. 
   
2. #include <iostream>
   
3. #include <pthread.h>
   
4. #include <sys/types.h>
   
5. #include <unistd.h>
   
6. using namespace std;
   
7. 
   
8. int number = 1;
   
9. 
   
10. void* thread1(void*)
    
11. {
    
12.   for(int i = 0; i < 10; i++)
    
13.   {
    
14.     cout << "this is thread1, number is " << number++;
    
15.     cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
    
16.   }
    
17. }
    
18. 
    
19. void* thread2(void*)
    
20. {
    
21.   for(int i = 0; i < 10; i++)
    
22.   {
    
23.     cout << "this is thread2, number is " << number++;
    
24.     cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
    
25.   }
    
26. }
    
27. 
    
28. int main()
    
29. {
    
30.   typedef void* (*thre)(void*);
    
31.   pthread_t tid[2];
    
32.   //thre thre1 = thread1;
    
33.   
    
34.   pthread_create(&tid[1], NULL, thread1, NULL);
    
35.   pthread_create(&tid[2], NULL, thread2, NULL);
    
36.   
    
37.   cout << "tid1 is " << tid[1] << " tid2 is " << tid[2] << endl;
    
38.   pthread_join(tid[1], NULL);
    
39.   pthread_join(tid[2], NULL);
    
40. 
    
41.   return 0;
    
42. }
    

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上面的程序可以正常编译,但是下面的就不行了

1. 
   
2. #include <iostream>
   
3. #include <pthread.h>
   
4. #include <sys/types.h>
   
5. #include <unistd.h>
   
6. using namespace std;
   
7. 
   
8. class Thread
   
9. {
   
10.   private:
    
11.     int number;
    
12.     pthread_t tid[2];
    
13. 
    
14.   public:
    
15.     Thread()
    
16.     {
    
17.       number = 1;
    
18.       pthread_create(&tid[1], NULL, thread1, NULL);
    
19.       pthread_create(&tid[2], NULL, thread2, NULL);
    
20.     }
    
21. 
    
22.     ~Thread()
    
23.     {
    
24.       cout << "tid1 is " << tid[1] << " tid2 is " << tid[2] << endl;
    
25.       pthread_join(tid[1], NULL);
    
26.       pthread_join(tid[2], NULL);
    
27.     }
    
28. 
    
29.     void* thread1(void*)
    
30.     {
    
31.       for(int i = 0; i < 10; i++)
    
32.       {
    
33.         cout << "this is thread1, number is " << number++;
    
34.         cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
    
35.       }
    
36.     }
    
37. 
    
38.     void* thread2(void*)
    
39.     {
    
40.       for(int i = 0; i < 10; i++)
    
41.       {
    
42.         cout << "this is thread2, number is " << number++;
    
43.         cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
    
44.       }
    
45.     }
    
46. };
    
47. 
    
48. int main()
    
49. {
    
50.   Thread *th = new Thread();
    
51.   delete(th);
    
52.   
    
53.   return 0;
    
54. }
    

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出现的错误是

1. 
   
2. thread.cpp: In constructor `Thread::Thread()':
   
3. thread.cpp:17: no matches converting function `thread1' to type `
   
4.    void*(*)(void*)'
   
5. thread.cpp:29: candidates are: void* Thread::thread1(void*)
   
6. thread.cpp:18: no matches converting function `thread2' to type `
   
7.    void*(*)(void*)'
   
8. thread.cpp:38: candidates are: void* Thread::thread2(void*)
   

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使用tepdef void*(*thre)(void*); thre ttt1 = thread1;作为参数还是不行
不晓得是甚么原因????
望高人指点

king_wuhan

似乎在那里看到，要把线程函数体定义在类外部，或者前加“static”

clyman

但是他报的是类型转换....

billzhou

原帖由 king_wuhan 于 2006-11-13 16:11 发表
似乎在那里看到，要把线程函数体定义在类外部，或者前加“static”

我试了一下  好象不太行啊

billzhou

线程函数访问私有变量,可以分两步实现
（1）函数声明为静态函数；
（2）在调用函数的时候将实例的指针this以void *的形式传递进去；

#include <iostream>
#include <pthread.h>
#include <sys/types.h>
#include <unistd.h>
using namespace std;

class Thread
{
  private:
    int number;
    pthread_t tid[2];

  public:
    Thread()
    {
      number = 1;
      pthread_create(&tid[1], NULL, thread1, this);
      pthread_create(&tid[2], NULL, thread2, this);
    }

    ~Thread()
    {
      cout << "tid1 is " << tid[1] << " tid2 is " << tid[2] << endl;
      pthread_join(tid[1], NULL);
      pthread_join(tid[2], NULL);
    }

   static void* thread1(void* arg)
    {
      Thread* nthis=(Thread *)arg;
      for(int i = 0; i < 10; i++)
      {
        cout << "this is thread1, number is " << number++;
        cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
      }
    }

   static void* thread2(void* arg)
    {
      Thread* nthis=(Thread *)arg;
      for(int i = 0; i < 10; i++)
      {
        cout << "this is thread2, number is " << number++;
        cout << " pid is " << getpid() << " id is " << pthread_self() << endl;
      }
    }
};

int main()
{
  Thread *th = new Thread();
  delete(th);
  
  return 0;
}

aloneme_live

原帖由 billzhou 于 2006-11-14 10:18 发表
线程函数访问私有变量,可以分两步实现
（1）函数声明为静态函数；
（2）在调用函数的时候将实例的指针this以void *的形式传递进去；

#include <iostream>
#include <pthread.h>
#include < ...

编译通不过啊，呵呵

billzhou

原帖由 billzhou 于 2006-11-14 10:18 发表
线程函数访问私有变量,可以分两步实现
（1）函数声明为静态函数；
（2）在调用函数的时候将实例的指针this以void *的形式传递进去；

#include <iostream>
#include <pthread.h>
#include < ...

忘了  函数体内改一下    nthis.number++就可以了

x.jc

把线程函数写成全局函数

converse

参见类成员函数的帖子,设置为static是对的

醉卧水云间

楼上两个对。