1. Си / Говнокод #29090

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    void main() {
    ph_fork f1, f2, f3, f4, f5;

    f_arr[0] = &f1;
    f_arr[1] = &f2;
    f_arr[2] = &f3;
    f_arr[3] = &f4;
    f_arr[4] = &f5;

    philosopher ph1, ph2, ph3, ph4, ph5;

    ph_arr[0] = &ph1;
    ph_arr[1] = &ph2;
    ph_arr[2] = &ph3;
    ph_arr[3] = &ph4;
    ph_arr[4] = &ph5;

    f1.number = 1;
    sem_init(&f1.is_free, 0, 1);

    f2.number = 2;
    sem_init(&f2.is_free, 0, 1);

    f3.number = 3;
    sem_init(&f3.is_free, 0, 1);

    f4.number = 4;
    sem_init(&f4.is_free, 0, 1);

    f5.number = 5;
    sem_init(&f5.is_free, 0, 1);

    ph1.number = 1;
    ph1.ph_fork_amount = 0;
    ph1.ph_forks[0] = 1;
    ph1.ph_forks[1] = 2;
    ph1.times_eaten = 0;
    ph1.times_thought = 0;
    ph1.st = THINKING;
    pthread_create(&ph1.thread, NULL, routine, (void*)&ph1);

    ph2.number = 2;
    ph2.ph_fork_amount = 0;
    ph2.ph_forks[0] = 2;
    ph2.ph_forks[1] = 3;
    ph2.times_eaten = 0;
    ph2.times_thought = 0;
    ph2.st = THINKING;
    pthread_create(&ph2.thread, NULL, routine, (void*)&ph2);

    ph3.number = 3;
    ph3.ph_fork_amount = 0;
    ph3.ph_forks[0] = 3;
    ph3.ph_forks[1] = 4;
    ph3.times_eaten = 0;
    ph3.times_thought = 0;
    ph3.st = THINKING;
    pthread_create(&ph3.thread, NULL, routine, (void*)&ph3);

    ph4.number = 4;
    ph4.ph_fork_amount = 0;
    ph4.ph_forks[0] = 4;
    ph4.ph_forks[1] = 5;
    ph4.times_eaten = 0;
    ph4.times_thought = 0;
    ph4.st = THINKING;
    pthread_create(&ph4.thread, NULL, routine, (void*)&ph4);

    ph5.number = 5;
    ph5.ph_fork_amount = 0;
    ph5.ph_forks[0] = 5;
    ph5.ph_forks[1] = 1;
    ph5.times_eaten = 0;
    ph5.times_thought = 0;
    ph5.st = THINKING;
    pthread_create(&ph5.thread, NULL, routine, (void*)&ph5);

    pthread_join(ph1.thread, NULL);
    pthread_join(ph2.thread, NULL);
    pthread_join(ph3.thread, NULL);
    pthread_join(ph4.thread, NULL);
    pthread_join(ph5.thread, NULL);

    sem_destroy(&f1.is_free);
    sem_destroy(&f2.is_free);
    sem_destroy(&f3.is_free);
    sem_destroy(&f4.is_free);
    sem_destroy(&f5.is_free);

    printf("\nThe lunch has ended!\n--------\nRESULTS:\nPhilosopher 1 has eaten %d times and thought %d times\nPhilosopher 2 has eaten %d times and thought %d times\nPhilosopher 3 has eaten %d times and thought %d times\nPhilosopher 4 has eaten %d times and thought %d times\nPhilosopher 5 has eaten %d times and thought %d times\n", ph1.times_eaten, ph1.times_thought, ph2.times_eaten, ph2.times_thought, ph3.times_eaten, ph3.times_thought, ph4.times_eaten, ph4.times_thought, ph5.times_eaten, ph5.times_thought);
}

    Решение задачи про обедающих философов, часть вторая.

    Запостил: GDMaster, 13 Февраля 2025

    Комментарии (1) RSS

    • ava guest6 9 дней назад # 0

      круто! а можете пожалуйста сделать вариант для сорока семи фолосов?
      Сейчас просто знаете бывает много ядер у процессоров
      Ответить

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