Grand Central Dispatch的简单例子

Question

我是mac的新手编程,我对Grand Central Dispatch感到非常惊讶.我读到了这个并且看起来像是并行编程的完美解决方案.我使用POSIX线程并希望转移到GCD.

我在Apple Developer Connection中看到了示例代码,但它让我很困惑.我搜索了一个简单的例子,有两个线程开始,但我找不到它.

我怎么能用GCD做这个示例代码???

#include <stdio.h>       /* standard I/O routines                 */
#include <pthread.h>     /* pthread functions and data structures */

/* function to be executed by the new thread */
void* do_loop(void* data)
{
int i;          /* counter, to print numbers */
int j;          /* counter, for delay        */
int me = *((int*)data);     /* thread identifying number */

for (i=0; i<10; i++) 
{
    for (j=0; j<500000; j++) /* delay loop */
    ;
    printf("'%d' - Got '%d'\n", me, i);
}

/* terminate the thread */
pthread_exit(NULL);
}
void* th2(void* data)
{
cout << "Thread nº 2" << endl;
}

int main(int argc, char* argv[])
{
int        thr_id;         /* thread ID for the newly created thread */
pthread_t  p_thread1;
pthread_t  p_thread2;       /* thread's structure                     */
int        a         = 1;  /* thread 1 identifying number            */
int        b         = 2;  /* thread 2 identifying number            */

/* create a new thread that will execute 'do_loop()' */
thr_id = pthread_create(&p_thread1, NULL, do_loop, (void*)&a);
/* run 'do_loop()' in the main thread as well */
thr_id = pthread_create(&p_thread2, NULL, th2, (void*)&b);


return 0;
}

提前致谢

Answer 1

它会是这样的:

// The block replaces your doLoop function, it basically does the same thing
dispatch_block_t myBlock = ^{
    int i;          /* counter, to print numbers */
    int j;          /* counter, for delay        */

    dispatch_queue_t me = dispatch_get_current_queue();     /* The queue which currently runs this block */

    for (i=0; i<10; i++) 
    {
        for (j=0; j<500000; j++) /* delay loop */
            ;


        printf("'%s' - Got '%d'\n", dispatch_queue_get_label(me), i); // Print the name of the queue
    }
};


// Create two queues
dispatch_queue_t queue1 = dispatch_queue_create("my.totally.unique.and.reverse.dns.identifier.1", NULL);
dispatch_queue_t queue2 = dispatch_queue_create("my.totally.unique.and.reverse.dns.identifier.2", NULL);

// Let both execute the block
dispatch_async(queue1, myBlock);
dispatch_async(queue2, myBlock);

// And then release the queues because we are great citizens who care about memory
dispatch_release(queue1);
dispatch_release(queue2);

Answer 2

只需将需要在传递给的块中异步运行的代码包装起来dispatch_async():

int main(int argc, char* argv[])
{
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0),
    ^ {
        // This code runs asynchronously!
        for (unsigned long long i = 0ULL; i < 100000000000ULL; i++) {
           cout << i << endl;
        }
    });

    return 0;
}

请注意,在此示例和您的示例中,应用程序实际上将立即返回,因为您不执行任何操作来推迟返回main().