Няшная / Говнокод #29089 Ссылка на оригинал

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#include <pthread.h>
#include <stdio.h>
#include <semaphore.h>
#include <unistd.h>
typedef struct _ph_fork {
    int number;
    sem_t is_free;
} ph_fork;
typedef enum state {
    EATING,
    THINKING
} state;
typedef struct _philosopher {
    int number;
    int ph_fork_amount;
    int ph_forks[2];
    int times_eaten;
    int times_thought;
    state st;
    pthread_t thread;
} philosopher;

ph_fork* f_arr[5] = {0};
philosopher* ph_arr[5] = {0};

void* routine(void* arg) {
    philosopher* ph = (philosopher*)arg;
    int frks[2] = {0};
    frks[0] = ph->ph_forks[0];
    frks[1] = ph->ph_forks[1];
    int ph_fork_left_odd = frks[0] % 2;
    int ph_fork_right_odd = frks[1] % 2;
    printf("Philosopher %d has taken a seat at the table.\n", ph->number);
    for (int i = 0; i < 72; i++) {
        if (!(ph_fork_left_odd && ph_fork_right_odd)) {
            if (ph_fork_left_odd) {
                sem_wait(&f_arr[frks[0] - 1]->is_free);
                printf("Philosopher %d takes left fork…\n", ph->number);
                ph->ph_fork_amount++;
                sem_wait(&f_arr[frks[1] - 1]->is_free);
                printf("Philosopher %d takes right fork…\n", ph->number);
                ph->ph_fork_amount++;
            }
            if (ph_fork_right_odd) {
                sem_wait(&f_arr[frks[1] - 1]->is_free);
                printf("Philosopher %d takes right fork…\n", ph->number);
                ph->ph_fork_amount++;
                sem_wait(&f_arr[frks[0] - 1]->is_free);
                printf("Philosopher %d takes left fork…\n", ph->number);
                ph->ph_fork_amount++;
            }
        } else { // Special case when both ph_forks are odd
            sem_wait(&f_arr[frks[0] - 1]->is_free);
            printf("Philosopher %d takes left fork…\n", ph->number);
            ph->ph_fork_amount++;
            sem_wait(&f_arr[frks[1] - 1]->is_free);
            printf("Philosopher %d takes right fork…\n", ph->number);
            ph->ph_fork_amount++;
        }
        if (ph->ph_fork_amount == 2) {
            ph->st = EATING;
            ph->times_eaten++;
            printf("Philosopher %d is eating…\n", ph->number);
            sleep(1);
        };
        if (!(ph_fork_left_odd && ph_fork_right_odd)) {
            if (ph_fork_left_odd) {
                printf("Philosopher %d puts back left fork…\n", ph->number);
                sem_post(&f_arr[frks[0] - 1]->is_free);
                ph->ph_fork_amount--;
                printf("Philosopher %d puts back right fork…\n", ph->number);
                sem_post(&f_arr[frks[1] - 1]->is_free);
                ph->ph_fork_amount--;
            }
            if (ph_fork_right_odd) {
                printf("Philosopher %d puts back right fork…\n", ph->number);
                sem_post(&f_arr[frks[1] - 1]->is_free);
                ph->ph_fork_amount--;
                printf("Philosopher %d puts back left fork…\n", ph->number);
                sem_post(&f_arr[frks[0] - 1]->is_free);
                ph->ph_fork_amount--;
            }
        } else { // Special case when both ph_forks are odd
            printf("Philosopher %d puts back left fork…\n", ph->number);
            sem_post(&f_arr[frks[0] - 1]->is_free);
            ph->ph_fork_amount--;
            printf("Philosopher %d puts back right fork…\n", ph->number);
            sem_post(&f_arr[frks[1] - 1]->is_free);
            ph->ph_fork_amount--;
        }
        if (ph->ph_fork_amount == 0) {
            ph->st = THINKING;
            printf("Philosopher %d is thinking…\n", ph->number);
            ph->times_thought++;
            sleep(1);
        }
    }
    printf("Philosopher %d has finished his lunch and left.\n", ph->number);
    return NULL;
};

Решение задачи про обедающих философов, часть первая.

Запостил: GDMaster GDMaster, (Updated )

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    А не использовать ли нам bbcode?


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