5

所以我将一组图像发布到我的服务器。我想使用 GCD 异步发布数组，但我也想让发生这种情况的方法同步，以便我可以传回单个响应对象。但是方法 dispatch_group_wait 似乎立即返回（而不是等待我的块完成）。这是一个问题，因为我在一个块中使用一个块吗？

NSArray *keys = [images allKeys];
__block NSMutableDictionary *responses = [NSMutableDictionary dictionaryWithCapacity:[images count]];

dispatch_group_t group = dispatch_group_create();
dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);
for (int i = 0; i < [keys count]; i++) {
    __block NSString *key = [keys objectAtIndex:i];
    dispatch_group_async(group, queue, ^{
        [self postImage:[images objectForKey:key] completionHandler:^(ServerResponse *response){
            @synchronized(responses){
                if ([response succeeded]) {
                    NSString *value = [[response data] objectForKey:@"image_token"];
                    [responses setObject:value forKey:key];
                    NSLog(@"inside success %@",responses);
                } else {
                    NSString *error = response.displayableError;
                    if (!error) {
                        error = @"Sorry something went wrong, please try again later.";
                    }
                    [responses setObject:error forKey:@"error"];
                    [responses setObject:response forKey:@"response"];
                }
            }
        }];
    });
}
dispatch_group_wait(group, DISPATCH_TIME_FOREVER);
dispatch_release(group);

我只想等待所有 [self postImage] 方法从服务器回调并修改响应字典。

4

3 回答 3

9

乔纳森的信号量示例是正确的。但是，我提到使用条件变量作为替代方案，所以我想我至少会发布一个示例。通常，CV 可用于等待更一般的条件，而不仅仅是 N 个工人。

请注意，条件变量有它们的位置（尽管不一定在这里），通常最好是在已经需要锁来改变共享状态时，然后其他线程可以等待特定条件。

NSUInteger numKeys = [keys count];

NSConditionLock *conditionLock = [[NSConditionLock alloc] initWithCondition:numKeys];
dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);
for (NSUInteger i = 0; i < numKeys; i++) {
    __block NSString *key = [keys objectAtIndex:i];
    dispatch_async(queue, ^{
        [self postImage:[images objectForKey:key] completionHandler:^(ServerResponse *response){
            // Basically, nothing more than a obtaining a lock
            // Use this as your synchronization primitive to serialize access
            // to the condition variable and also can double as primitive to replace
            // @synchronize -- if you feel that is still necessary
            [conditionLock lock];

            ...;

            // When unlocking, decrement the condition counter
            [conditionLock unlockWithCondition:[conditionLock condition]-1];
        }];
    });
}

// This call will get the lock when the condition variable is equal to 0
[conditionLock lockWhenCondition:0];
// You have mutex access to the shared stuff... but you are the only one
// running, so can just immediately release...
[conditionLock unlock];
于 2012-08-19T23:39:44.887 回答
7

没有看到 的代码-postImage:completionHandler:，很难说事情是在哪里安排的，但我假设他们调用了 iOS 提供的东西。如果是这样，块内的处理程序块将异步分派到全局队列，然后 iOS 提供的函数或方法会立即返回。就您的调度组而言，工作几乎是立即完成的。

没有简单的方法可以让团队等待在调用时间之前尚未安排好的工作dispatch_group_wait()。但是，我们可以添加一个称为信号量的较低级别的thingamajigger ，以确保我们的操作以正确的顺序完成，并将其安排在内部（异步）块的范围之外。

NSUInteger numKeys = [keys count];

dispatch_group_t group = dispatch_group_create();
dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);
for (NSUInteger i = 0; i < numKeys; i++) {
    __block NSString *key = [keys objectAtIndex:i];
    dispatch_group_async(group, queue, ^{
        // We create a semaphore for each block here. More on that in a moment.
        // The initial count of the semaphore is 1, meaning that a signal must happen
        // before a wait will return.
        dispatch_semaphore_t sem = dispatch_semaphore_create(1);
        dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
        [self postImage:[images objectForKey:key] completionHandler:^(ServerResponse *response){
            ...
            // This instructs one caller (i.e. the outer block) waiting on this semaphore to resume.
            dispatch_semaphore_signal(sem);
        }];

        // This instructs the block to wait until signalled (i.e. at the end of the inner block.)
        dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);

        // Done with the semaphore. Nothing special here.
        dispatch_release(sem);
    });
}
dispatch_group_wait(group, DISPATCH_TIME_FOREVER);
// Now all the tasks should have completed.
dispatch_release(group);

不过这里有个问题。信号量是一种内核资源。如果我们有 100 个任务要执行，但内核只能提供 99 个信号量怎么办？坏事™发生。我们可以重建代码以仅使用一个信号量，尽管等待它会显得有点不靠谱。顺便说一句，这样做实际上完全消除了调度组，所以我们实际上是用信号量替换了组。让我们来看吧！

NSUInteger numKeys = [keys count];

// set the count of the semaphore to the number of times it must be signalled before
// being exhausted. Up to `numKeys` waits will actually wait for signals this way.
// Additional waits will return immediately.
dispatch_semaphore_t sem = dispatch_semaphore_create(numKeys);
for (int i = 0; i < numKeys; i++) {
    dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
}

dispatch_queue_t queue = dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0);
for (NSUInteger i = 0; i < numKeys; i++) {
    __block NSString *key = [keys objectAtIndex:i];
    dispatch_async(queue, ^{
        [self postImage:[images objectForKey:key] completionHandler:^(ServerResponse *response){
            ...;

            // This decrements the semaphore's count by one. The calling code will be
            // woken up by this, and will then wait again until no blocks remain to wait for.
            dispatch_semaphore_signal(sem);
        }];
    });
}

// At this point, all the work is running (or could have already completed, who knows?).
// We don't want this function to continue running until we know all of the blocks
// have run, so we wait on our semaphore a number of times equalling the number
// of signals we expect to get. If all the blocks have run to completion before we've
// waited for all of them, the additional waits will return immediately.
for (int i = 0; i < numKeys; i++) {
    dispatch_semaphore_wait(sem, DISPATCH_TIME_FOREVER);
}
// Now all the tasks should have completed.
dispatch_release(sem);
于 2012-08-19T22:32:20.703 回答
3

是的，如前所述，dispatch_group_wait()不等待，因为postImage:completionHandler:调用似乎是异步的。如果你真的需要这段代码同步执行，那么信号量或锁似乎是一个合适的解决方案。

但是，如果您只想将所有响应收集到一本字典中进行处理，我相信最合适的解决方案是充分利用 GCD。并使用调度队列来管理可变字典；在我看到的大多数文件中，这似乎是苹果首选的解决方案。

解决方案的关键是本质上将可变字典的所有权转移到单个队列，然后仅从该队列中修改它。我所指的“所有权”不是内存管理意义上的对象所有权，而是修改权意义上的所有权。

我会考虑做这样的事情：

NSArray *keys = [images allKeys];
// We will not be reasigning the 'responses' pointer just sending messages to the NSMutableDictionary object __block is not needed.
NSMutableDictionary *responses = [NSMutableDictionary dictionaryWithCapacity:[images count]];
// Create a queue to handle messages to the responses dictionary since mutables are not thread safe.
// Consider this thread the owner of the dictionary.
dispatch_queue_t responsesCallbackQueue = dispatch_queue_create("com.mydomain.queue", DISPATCH_QUEUE_SERIAL);

for (NSString *key in keys) {
    // This async call may not be needed since postImage:completionHandler: seems to be an async call itself.
    dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{ 
        [self postImage:[images objectForKey:key] completionHandler:^(ServerResponse *response){
            dispatch_async(responsesCallbackQueue, ^{
                [responses setObject:response forKey:key];
                // Perhaps log success for individual requests.
                if (responses.count == keys.count){
                    NSLog(@"All requests have completed");
                    // All responses are available to you here.
                    // You can now either transfer back 'ownership' of the dictionary.
                    dispatch_async(dispatch_get_main_queue(), ^{
                        [self requestsHaveFinished:responses];
                    });
                }
            });
        }];
    });
}
于 2012-08-20T00:44:01.917 回答