面试题-多线程-如何在N个任务都完成后才执行A任务

方法一：使用dispatch_group

dispatch_group_t group = dispatch_group_create();
    for (int i=0; i<10; i++) {
        dispatch_group_enter(group);
        dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
            sleep(1);
            printf("i = %d", i);
            dispatch_group_leave(group);
        });
    }
dispatch_group_notify(group, dispatch_get_main_queue(), ^{
        printf("开始执行任务A");
    });

方法二： 信号量实现

let sem = DispatchSemaphore.init(value: 0)
        for i in 0..<10 {
            DispatchQueue.global().async {
                sleep(UInt32(1))
                print("\(i)")
                sem.signal()
            }
            
        }
        for _ in 0..<10 {
            sem.wait()
        }
        DispatchQueue.main.async {
            print("开始执行任务A")
        }

方法三： 简单粗暴的bool数组，不推荐, 高并发容易崩溃

    self.arr = [[NSMutableArray alloc] init];
    for (int i=0; i<10; i++)
    {
        [self.arr addObject: [NSNumber numberWithBool:false]];
    }
    for (int i=0; i<10; i++) {
        dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), ^{
            sleep(i);
            self.arr[i] = [[NSNumber alloc] initWithBool:true];
            if (![self.arr containsObject:[NSNumber numberWithBool:false]]) {
                NSLog(@"开始执行任务A");
            }
        });
    }

如果将sleep(i)改成sleep(1)就会发现崩溃了，原因是因为在遍历arr的时候对arr进行了修改。