#include <pthread.h>
#include <semaphore.h>
#include <stdio.h>

/*
This program provides a possible solution for the first readers-writers problem
using a mutex and a semaphore.

10 readers and 5 writers are used to demonstrate the solution. Feel free to
change these values.
*/

sem_t wrt;
pthread_mutex_t mutex;
int cnt = 1;
int numreader = 0;

void *writer(void *wno)
{
    sem_wait(&wrt);
    cnt = cnt * 2;
    printf("Writer %d modified cnt to %d\n", *((int *)wno), cnt);
    sem_post(&wrt);
    return NULL;
}

void *reader(void *rno)
{
    // Reader acquires the lock before modifying numreader
    pthread_mutex_lock(&mutex);
    numreader++;
    if (numreader == 1) {
        sem_wait(&wrt); // If this is the first reader, it blocks the writer
    }
    pthread_mutex_unlock(&mutex);

    // Reading section
    printf("Reader %d: read cnt as %d\n", *((int *)rno), cnt);

    // Reader acquires the lock before modifying numreader
    pthread_mutex_lock(&mutex);
    numreader--;
    if (numreader == 0) {
        sem_post(&wrt); // If this is the last reader, it wakes up the writer
    }
    pthread_mutex_unlock(&mutex);

    return NULL;
}

int main()
{
    pthread_t read_t[10], write_t[5];
    pthread_mutex_init(&mutex, NULL);
    sem_init(&wrt, 0, 1);

    int a[10] = {1,2,3,4,5,6,7,8,9,10}; // Just used for numbering readers/writers

    for (int i = 0; i < 10; i++) {
        pthread_create(&read_t[i], NULL, reader, (void *)&a[i]);
    }

    for (int i = 0; i < 5; i++) {
        pthread_create(&write_t[i], NULL, writer, (void *)&a[i]);
    }

    for (int i = 0; i < 10; i++) {
        pthread_join(read_t[i], NULL);
    }

    for (int i = 0; i < 5; i++) {
        pthread_join(write_t[i], NULL);
    }

    pthread_mutex_destroy(&mutex);
    sem_destroy(&wrt);

    return 0;
}