Indian – The name on the ticket should be according to passport or visa

My cousin is planning to visit me in the US and has approved a B1 / B2 visa. Her name on the pass is: FIRST NAME – AAA, LAST NAME – BLANK, while the name on her visa reads: FIRST NAME – FNU, LAST NAME – AAA. Now, when she books her ticket from India to the US, what name should she use? Should it be loud passport or visa? She plans to travel with Air India. When she leaves India, the airport authorities check her passport data, in the United States the immigration officers check her visa. So we do not know which name pattern to follow, and book their tickets accordingly. We are really confused! Is anyone else in the same situation?

air travel – skip leg, let ticket expire – mistakes happen due to late connections?

As we know, airlines fight ticketing in hidden cities by canceling your entire ticket (including return) if you miss a leg. But how is this handled – what happens if you are too late from your connection, this can lead to misfires? I think the question is, do you have to worry that a late connection will cause your entire ticket to expire? Most of all, the delay on trans-oceanic routes, which run only once a day, can easily be a day or worse at worst.

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c ++ multithreading ticket class to wait for completion of parallel tasks

I have implemented a "ticket" class that is shared between threads as shared_ptr.

The program flow is as follows:

  1. parallelQuery () is called to start a new query job. A shared instance of Ticket is created.
  2. The query is divided into several tasks. Each task is queued in a worker thread. Each task receives the common ticket.
  3. ticket.wait () is called to wait for all tasks in the job to complete.
  4. When a task is done, the done () method for the ticket is called.
  5. When all tasks have been completed, the ticket is unlocked. The result data of the task is aggregated and returned by parallelQuery ().

In pseudocode:

     std::vector parallelQuery(std::string str) {
         auto ticket = std::make_shared();
         auto task1 = std::make_unique(ticket, str+"a");
         addTaskToWorker(task1);
         auto task2 = std::make_unique(ticket, str+"b");
         addTaskToWorker(task2);
         ticket->waitUntilDone();
         auto result = aggregateData(task1, task2);
         return result;
     }

My code works. However, it is a critical part of my application and I would like to have a second opinion, especially on blocking and unlocking when the task is completed.

Here is the ticket class:

#include 
#include 

    class Ticket {
    public:
        Ticket(int numTasks = 1) : _numTasks(numTasks), _done(0), _canceled(false) {
            _mutex.lock();
        }

        void waitUntilDone() {
            _doneLock.lock();
            if (_done != _numTasks) {
                _doneLock.unlock();
                _mutex.lock();
            }
            else {
                _doneLock.unlock();
            }
        }

        void done() {
            _doneLock.lock();
            _done++;
            if (_done == _numTasks) {
                _mutex.unlock();
            }
            _doneLock.unlock();
        }

        void cancel() {
            _canceled = true;
            _mutex.unlock();
        }

        bool wasCanceled() {
            return _canceled;
        }

        bool isDone() {
            return _done >= _numTasks;
        }

        int getNumTasks() {
            return _numTasks;
        }

    private:
        std::atomic _numTasks;
        std::atomic _done;
        std::atomic _canceled;
        // mutex used for caller wait state
        std::mutex _mutex;
        // mutex used to safeguard done counter with lock condition in waitUntilDone
        std::mutex _doneLock;
    };