20220906c2

In questa pagina sono presenti alcune soluzioni (possibilmente con errore) per l'esercizio C2 dell'esame 2022/09/06. Il lettore è invitato a immedesimare il professore al momento della correzione durante la lettura delle soluzioni.

1

semaphore sem(1);
int sum=0;
queue q;

void sumstop(int v):
	if i < 0:
		q.enqueue(v)
		sem.P()
	else
		i--

int sumgo(void):
	for each elem in q:
		int val=q.dequeue()
		sum=sum + val
	return sum

2

queue values;
semaphore s(0);
int sum=0;

void sumstop(int v):
	s.P()
	values.enqueue(v)

int sumgo(void):
	while !value.isEmpty():
		s.V()
		mutex.P()
		sum=sum+values.dequeue()
		mutex.v()
	return sum

3

shared int sum=0;
semaphore wait(0);
semaphore mutex(1);

void sumstop(int v):
	mutex.P();
	sum = sum + v;
	mutex.V();
	wait.P();

int sumgo(void):
	wait.V();
	somma = sum;
	mutex.P()
	sum = 0
	mutex.V()
	return somma;

4

queue<int> processlist;
semaphore mutex(1);
semaphore s(0);

void sumstop(int v):
	mutex.P();
	processlist.enqueue(v);
	mutex.P();
	S.P();

int sumgo(void):
	int sum = 0;
	mutex.P();
	while processlist.top() != NULL:
		sum += processlist.dequeue();
		S.V();
	mutex.V()
	return sum;

5

int counter = 0;
int sum = 0;
semaphore mutex(1);
semaphore blocked(0);

void sumstop(int v):
	mutex.P();
	sum+=v;
	counter++;
	mutex.V();
	blocked.P();

int sumgo(void):
	  mutex.P();
		intret = sum;
		sum = 0;
		for (int i=0; i<counter; i++)
			blocked.V()
		counter = 0;
		mutex.V()
	return intret

6

int sum = 0;
mutex = new semaphore(1);
queue q;

void sumstop(int v):
	mutex.P();
	sum += v;
	sintpid = getpid();
	suspend(pid);
	mutex.V();

int sumgo(void):
	mutex.P();
	while !q.empty():
		resume(q.dequeue())
	int value = sum;
	sum = 0;
	return value;

7

int nP, nV = -1, 0;
int somma = 0
int val[]
semaphore mutex(1);
semaphore s(0);

void sumstop(int v):
	mutex.P()
	val[nP++].add[v]
	s.P()
	np--
	v[0].delete

int sumgo(void):
	mutex.P();
	nV++
	if nP > 0:
		s.V()
		for (int i=0; i<nV; i++)
			somma += val[i]
			nv--
	else
		mutex.V()
	return somma

8

struct blocked {
	semaphore sem(0);
	int value = 0;
};

semaphore mutex(1);
list<blocked> procs = new list<blocked>();

void sumstop(int v):
	mutex.P()
	blocked bl = new blocked();
	bl.value = v
	procs.add(bl)
	bl.sem.P()
	mutex.V()

int sumgo(void):
	mutex.P();
	int count = 0;
	foreach proc in procs:
		count += proc.value
		procs.remove(proc)
		proc.sem.V()
	mutex.V()
	return count;

== 9==

semaphore s[] new sem(0)
semaphore s1 new sem(0)
int tot, waiting = 0;

void sumstop(int v):
	waiting++
	s[waiting - 1].P()
	waiting--
	tot = tot + v
	if waiting == 0:
		s1.V()

int sumgo(void):
	if waiting = 0:
		return 0
	for (i = waiting -1; i == 0; i--)
		s[i].V()
	s1.P() // per aspettare che tutti abbiano fatto la somma
	return tot

10

int nw = 0
int currsum = 0
semaphore mutex(1)
semaphore wait2go(0)

void sumstop(int v):
	mutex.P()
	currsum += v;
	nw++;
	mutex.V()
	wait2go.P()
	if --nw > 0:
		wait2go.V()
	else
		mutex.V()

int sumgo(void):
	mutex.P();
	if nw == 0:
		mutex.V()
		return 0;
	int sum = cursum;
	cursum = 0;
	wait2go.V();
	return sum

11

semaphore mutex(1);
semaphore semwait(0);
int sum=0;
int wait=0;

void sumstop(int v):
	mutex.P()
	wait++
	sum += v;
	mutex.V()
	semwait.P()
	mutex.P()
	wait--;
	mutex.V()
	if wait > 0:
		semwait.V()

int sumgo(void):
	if wait == 0:
		return 0
	semwait.V();
	while (wait > 0) {}
	int val = sum
	mutex.P()
	sum = 0;
	mutex.V()
	return val;

12

semaphore mutex(0)
volatile int counter = 0

void sumstop(int v):
	counter = counter + v;
	mutex.P()

int sumgo(void):
	int val = counter;
	while (mutex.value != 0)
		mutex.V()
	counter = 0;
	return counter

13

semaphore mutex(1)
int sum = 0;
queue of semaphore q;

void sumstop(int v):
	mutex.P()
	sum += v;
	s = new semaphore(0);
	q.enqueue(s)
	mutex.V()
	s.P()
	free(s)

int sumgo(void):
	mutex.P()
	int lsum = sum
	sum = 0
	while (!q.empty()):
		semaphore s = q.dequeue()
		s.V()
	mutex.V()

14

semaphore mutex(1)
int sum = 0;
queue of semaphore q;

void sumstop(int v):
	mutex.P()
	sum += v;
	s = new semaphore(0);
	q.enqueue(s)
	mutex.V()
	s.P()
	free(s)
	if(q.empty())
		mutex.V()
	else
		semaphore s = q.dequeue()
		s.V()

int sumgo(void):
	mutex.P()
	int lsum = sum
	sum = 0
	if(q.empty())
		mutex.V()
	else
		semaphore s = q.dequeue()
		s.V()

15 (soluzione proposta su telegram)

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

semaphore s; // semaforo che blocca le somme
semaphore finished; // semaforo che indica che le somme sono state fatte
semaphore critical_section; // sarà utilizzata per decidere chi entra nella critical section

volatile int sumstops = 0;
volatile int result;

void *sumstop(int v) {
    semaphore_P(critical_section);
    sumstops++;
    semaphore_V(critical_section);

    semaphore_P(s);
    printf("summed %d\n", v);
    result += v;
    semaphore_V(finished);
}


int sumgo(void) {
    semaphore_P(critical_section); // così sumstops è costante all'interno di questa section
    result = 0;
    for (int i = 0; i < sumstops; i++) {
        semaphore_V(s); // permetti alla somma di andare
        semaphore_P(finished); // aspetta che la somma sia finita prima di continuare
    }
    sumstops = 0; // reset number of blocked stops.
    semaphore_V(critical_section);
    return result;
}

void *run_sumgo(void) {
    int res = sumgo();
    printf("the result found is %d\n", res);
    printf("the result found is %d\n", result);
}

int main() {
    srand(time(NULL));
    s = semaphore_create(0);
    critical_section = semaphore_create(1);
    finished = semaphore_create(0);

    int n = rand() % 50;
    pthread_t sumg, s[n];

    for (int i = 0; i < n; i++) {
        pthread_create(&s[i], NULL, sumstop, i);
    }
    pthread_create(&sumg, NULL, run_sumgo, NULL);

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

    pthread_join(sumg, NULL);
}