C-带循环的多个子叉

Question

So I have to create 7 child processes with fork, and each one has to work in loop until medicamentos (meds) runs out. But they only run one at a time, and sometimes only once. The variables and other names are in pt-br, but I'll put translations. So, it was supposed to be an nursing home (a terrible one tho heheheh) and it has a limited amount of meds. A person needs 1 pill to sleep for 4 hours and 8 for 8 hours. Also, there's only 5 beds, resulting on a "waiting line" for the beds. Once there's a free bed, an elder that's awake and waiting for a bed will take 1 or 2 pills, at random and sleep. After that, he'll leave the bed, letting another elder sleep.

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <unistd.h>

#define ACORDADO 0 //awake
#define ESPERANDO 1 //waiting
#define DORMINDO 2 //sleeping
#define N 7

void mostrar(void); //show info
void dormir(int); //sleep
void tomar_med(int); //take meds
void esperar(int);//wait
void acordar(int);//wake up
void liberar_cama(int);//leave the bed
void acao_idoso(int j);//elder action

int medicamentos = 90; //meds
int camas = 5; //beds
int idosos [N]; //elders
int i;

void forker(int n_proc)
{
    pid_t pid;

    if(n_proc > 0)
    {
        if ((pid = fork()) < 0){
            perror("fork");

        }else if (pid == 0){ //filho
            acao_idoso(n_proc - 1);

        }else if(pid > 0){ //pai
            mostrar();
            forker(n_proc - 1);
        }
    }
}

void main(){


    forker(7);

}

void mostrar(){
    for(i=0;i<N;i++){
        if(idosos[i] == ACORDADO)
            printf("O idoso %d esta acordado.\n", i+1); //elder is awake
        if(idosos[i] == ESPERANDO)
            printf("O idoso %d esta esperando.\n", i+1); //elder is waiting
        if(idosos[i] == DORMINDO)
            printf("O idoso %d esta dormindo.\n", i+1); //elder is asleep
    }
    printf("\n");   
    printf("%d medicamentos sobrando.\n", medicamentos); //%d meds left
    printf("\n");
}

void acao_idoso(int j){
    int i = j;
    while(medicamentos != 0){
        acordar(i);
        mostrar();
        esperar(i);
        mostrar();
        dormir(i);
        mostrar();
        liberar_cama(i);
        mostrar();
    }
}

void acordar(int i){
    if(idosos[i] == ACORDADO){
        int t = rand() %11;
        sleep(t+1);
    }
}

void esperar(int i){
    if(idosos[i] == ACORDADO && medicamentos > 0)
        idosos[i] = ESPERANDO;
}

void dormir(int i){
    if((idosos[i] == ESPERANDO) && (camas >= 1)){
        int j = rand();

        if(i % 2 == 0){
            medicamentos = medicamentos - 2;
            sleep(8);
        }else{
            medicamentos = medicamentos - 1;
            sleep(4);
        }

        camas = camas - 1;
        idosos[i] = DORMINDO;

    }
}

void liberar_cama(int i){
    if(idosos[i] == DORMINDO){
        idosos[i] = ACORDADO;
        camas++;
    }
}

Answer 1

As noted in the comments, your main program needs to wait for the children to finish, and you need to use j in dormir() .

If you sleep 4, 8, more seconds at a time, it will take a long time (10+ minutes) to complete. Change the sleep calls to sleep10(int n) to sleep n tenths of a second, and providing the implementation using nanosleep() , the code completes in something like 75 seconds.

You also need to ensure that no-one takes 2 pills when there's only one pill left; getting a negative 'medicamentos sobrando' leaves the program running indefinitely.

On Linux, void main() is unconditionally wrong; use int main(void) and (preferably) an explicit return 0; at the end of main() unless you have an alternative status to return.

The code also prints out PID numbers before messages to identify which process is doing what. There is room to think that there should be more information printed, but my Spanish is sufficiently non-existent that I'm not going to attempt it. The main() program reports on the children and their exit statuses as they exit.

#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/wait.h>
#include <unistd.h>
#include <time.h>

#define ACORDADO 0
#define ESPERANDO 1
#define DORMINDO 2
#define N 7

void mostrar(void);
void dormir(int);
void tomar_med(int);
void esperar(int);
void acordar(int);
void liberar_cama(int);
void acao_idoso(int j);
void forker(int n_proc);

int medicamentos = 90;
int camas = 5;
int idosos[N];

int main(void)
{
    forker(N);
    int corpse;
    int status;
    while ((corpse = wait(&status)) > 0)
        printf("PID %5d exited with status 0x%.4X\n", corpse, status);
    return 0;
}

static void sleep10(int n)
{
    assert(n > 0 && n < 100);
    struct timespec ts = { .tv_sec = n / 10, .tv_nsec = (n % 10) * 100000000 };
    printf("%d: sleeping %d.%ds\n", (int)getpid(), n / 10, n % 10);
    nanosleep(&ts, 0);
}

void forker(int n_proc)
{
    pid_t pid;

    if (n_proc > 0)
    {
        if ((pid = fork()) < 0)
        {
            perror("fork");
        }
        else if (pid == 0)
        {
            printf("%d: PID %d\n", n_proc, (int)getpid());
            acao_idoso(n_proc - 1);
        }
        else if (pid > 0)
        {
            mostrar();
            forker(n_proc - 1);
        }
    }
}

void mostrar()
{
    int pid = getpid();
    for (int i = 0; i < N; i++)
    {
        if (idosos[i] == ACORDADO)
            printf("%d: O idoso %d esta acordado.\n", pid, i + 1);
        if (idosos[i] == ESPERANDO)
            printf("%d: O idoso %d esta esperando.\n", pid, i + 1);
        if (idosos[i] == DORMINDO)
            printf("%d: O idoso %d esta dormindo.\n", pid, i + 1);
    }
    printf("\n");
    printf("%d: %d medicamentos sobrando.\n", pid, medicamentos);
    printf("\n");
}

void acao_idoso(int j)
{
    int i = j;
    while (medicamentos != 0)
    {
        acordar(i);
        mostrar();
        esperar(i);
        mostrar();
        dormir(i);
        mostrar();
        liberar_cama(i);
        mostrar();
    }
}

void acordar(int i)
{
    if (idosos[i] == ACORDADO)
    {
        int t = rand() % 11;
        sleep10(t + 1);
    }
}

void esperar(int i)
{
    if (idosos[i] == ACORDADO && medicamentos > 0)
        idosos[i] = ESPERANDO;
}

static inline int min(int x, int y) { return (x < y) ? x : y; }

void dormir(int i)
{
    if ((idosos[i] == ESPERANDO) && (camas >= 1))
    {
        int j = rand();

        if (j % 2 == 0)
        {
            medicamentos = medicamentos - min(2, medicamentos);
            sleep10(8);
        }
        else
        {
            medicamentos = medicamentos - 1;
            sleep10(4);
        }

        camas = camas - 1;
        idosos[i] = DORMINDO;
    }
}

void liberar_cama(int i)
{
    if (idosos[i] == DORMINDO)
    {
        idosos[i] = ACORDADO;
        camas++;
    }
}