Question

struct node
{
public:
    char *s;
    int up;
    node()
    {
        up = 0;
        s = new char[1000];
        memset (s, 0, sizeof(char) * 1000);
    }
    ~node()
    {
        delete [] s;
    }
    void insert()
    {
        s[up++] = 'a';
    }
};

void* test_thread(void *arg)
{
    pthread_mutex_lock( &mutex1 );
    node n;
    n.insert();
    printf ("%s\n", n.s);
    printf ("%x\n", &n);
    pthread_mutex_unlock( &mutex1 );
    pthread_exit(0);
    //return 0;
}

supose this function will be executed by

pthread_create(&id1, NULL, test_thread, NULL);
pthread_create(&id2, NULL, test_thread, NULL);

and it is compiled by

g++ test_thread.cpp -o main -lpthread -g 

its result is

a
40a001a0
a
40a001a0

In my Linux operator ,the address of node n in the two thread are the same!

I want to know why the address of node n in which the tho thread contains are the same?

Any answer is appreciated~~~

Thanks~~~

Answer 1

Add sleep(1) before you exit the thread. Now you should see two different addresses but the same output of 'a'. (you would need pthread_join though).

Now if you want to print 'aa' then you might have to define node in global space or define it in main.

With your current code the lock/unlock does not have any use but once you use the shared memory the 2nd thread can not write until the 1st thread has finished.