Sistemi Operativi - User contributions [en]

Prova teorica 2015.02.14

2017-06-06T10:08:40Z

MarcoNegrini: /* Soluzione di MarcoNegrini */

Testo: [http://www.cs.unibo.it/~renzo/so/compiti/2015.02.14.tot.pdf]

== Esercizio c.1 ==
===Soluzione di Silas===
<syntaxhighlight lang="C">
#define MAX

monitor altcolbb{
queue buff;
color_t last_color; //0: red, 1:blue, -1:"superstate", means that both red and blue are accepted
int wait_red, wait_blue;
condition ok2read, redok, blueok;

void altcolbb(void){
buff = new queue();
last_color = -1;
wait_red = wait_blue = 0;
}

procedure entry void write(color_t color, generic_type val){
if(last_color == color || buff.length() == MAX){ //we can't enqueue if the colors are the same or if the buffer is full
if(color == 0){ //enqueue to "reds"
wait_red++;
redok.wait();
wait_red--;
}else{ //enqueue to "blues"
wait_blue++;
blueok.wait();
wait_blue--;
}
}
buff.enqueue(val); //append val and update last_color
last_color = color;
ok2read.signal();
}

procedure entry generic_type read(void){
if(buff.length() == 0)
ok2read.wait();
generic_type ret = buff.dequeue();
if(buff.length() == 0)
last_color = -1; //if buff is empty both colors can now be added
switch(last_color){
case -1: {
if(wait_red>wait_blue) //if there are more "reds" waiting to write we signal them
redok.signal();
else
blueok.signal(); //otherwise we signal "blues"
break;
}
case 0: {
blueok.signal();
break;
}
case 1: {
redok.signal();
break;
}
}
}

}
</syntaxhighlight>

===Soluzione di MarcoNegrini===
I corrected Silas's version
<syntaxhighlight lang="C">

#define MAX

monitor altcolbb{
queue buff;
color_t last_color; //0: red, 1:blue, -1:"superstate", means that both red and blue are accepted
condition ok2read, ok2write, redok, blueok;

void altcolbb(void){
buff = new queue();
last_color = -1;
}

procedure entry void write(color_t color, generic_type val){
if(buff.length() == MAX){ //here waits if buff is full
ok2write.wait();
if(last_color==-1){ //if buff was empty it can just enqueue and exit
buff.enqueue(val);
last_color = color;
}
else{
if(color == 0){ //red section
if (last_color == 0) //if last was red it needs to wait
redok.wait();
buff.enqueue(val);
last_color = color; // it must set last color before signaling other writers
blueok.signal(); //blue can now enqueue his value
}else{ //blue section, same as above
if (last_color == 1)
blueok.wait();
buff.enqueue(val);
last_color = color;
redok.signal();
}
}
ok2read.signal(); // reader can now read
}

procedure entry generic_type read(void){
//here waits if needed, then dequeue
if(buff.length() == 0)
ok2read.wait();
generic_type ret = buff.dequeue();

//here signal the one that has been waiting for more time
// it MUST be of the last_color color
if(buff.length() == 0){
color_t tmp=last_color;
last_color = -1;
// last color MUST be set to -1 here because
// if there were writers waiting on ok2write.wait() (full buffer case)
// they woundn't be signaled here, this happens if MAX=1

if (tmp == 1) //
blueok.signal();
else
redok.signal();
}
ok2write.signal();
}

}
</syntaxhighlight>

== Esercizio c.2 ==
===Soluzione di Silas===
I semafori garantiscono che A e B vengano eseguiti in maniera sequenziale e che accedano in maniera mutualmente esclusiva ad n, quindi l'unica variazione possibile è il thread di inizio per la sequenza.
Nel caso sia A ad accedere per primo alla CS si ha: n = ((1*2)+2)*3 = 12
Nel caso sia invece B ad accedere per primo alla C si ha: n = (((0*2)+1)*3)+2 = 5
Quindi n = 12 e n = 5 sono i due possibili valori.

Prova teorica 2015.02.14

2017-06-06T10:07:22Z

MarcoNegrini:

Testo: [http://www.cs.unibo.it/~renzo/so/compiti/2015.02.14.tot.pdf]

== Esercizio c.1 ==
===Soluzione di Silas===
<syntaxhighlight lang="C">
#define MAX

monitor altcolbb{
queue buff;
color_t last_color; //0: red, 1:blue, -1:"superstate", means that both red and blue are accepted
int wait_red, wait_blue;
condition ok2read, redok, blueok;

void altcolbb(void){
buff = new queue();
last_color = -1;
wait_red = wait_blue = 0;
}

procedure entry void write(color_t color, generic_type val){
if(last_color == color || buff.length() == MAX){ //we can't enqueue if the colors are the same or if the buffer is full
if(color == 0){ //enqueue to "reds"
wait_red++;
redok.wait();
wait_red--;
}else{ //enqueue to "blues"
wait_blue++;
blueok.wait();
wait_blue--;
}
}
buff.enqueue(val); //append val and update last_color
last_color = color;
ok2read.signal();
}

procedure entry generic_type read(void){
if(buff.length() == 0)
ok2read.wait();
generic_type ret = buff.dequeue();
if(buff.length() == 0)
last_color = -1; //if buff is empty both colors can now be added
switch(last_color){
case -1: {
if(wait_red>wait_blue) //if there are more "reds" waiting to write we signal them
redok.signal();
else
blueok.signal(); //otherwise we signal "blues"
break;
}
case 0: {
blueok.signal();
break;
}
case 1: {
redok.signal();
break;
}
}
}

}
</syntaxhighlight>

===Soluzione di MarcoNegrini===
I corrected Silas's version
<syntaxhighlight lang="C">

#define MAX

monitor altcolbb{
queue buff;
color_t last_color; //0: red, 1:blue, -1:"superstate", means that both red and blue are accepted
condition ok2read, ok2write, redok, blueok;

void altcolbb(void){
buff = new queue();
last_color = -1;
}

procedure entry void write(color_t color, generic_type val){
if(buff.length() == MAX){ //here waits if buff is full
ok2write.wait();
if(last_color==-1){ //if buff was empty it can just enqueue and exit
buff.enqueue(val);
last_color = color;
}
else{
if(color == 0){ //red section
if (last_color == 0) //if last was red it needs to wait
redok.wait();
buff.enqueue(val);
last_color = color; // it must set last color before signaling other writers
blueok.signal(); //blue can now enqueue his value
}else{ //blue section, same as above
if (last_color == 1)
blueok.wait();
buff.enqueue(val);
last_color = color;
redok.signal();
}
}
ok2read.signal(); // reader can now read
}

procedure entry generic_type read(void){
//here waits if needed, then dequeue
if(buff.length() == 0)
ok2read.wait();
generic_type ret = buff.dequeue();

//here signal the one that has been waiting for more time
// it MUST be of the last_color color
if(buff.length() == 0){
color_t tmp=last_color;
last_color = -1;
// last color MUST be set to -1 here because
// if there was writers waiting on ok2write.wait() (full buffer case)
// they woundn't be signaled here, this happens if MAX=1

if (tmp == 1) //
blueok.signal();
else
redok.signal();
}
ok2write.signal();
}

}
</syntaxhighlight>

== Esercizio c.2 ==
===Soluzione di Silas===
I semafori garantiscono che A e B vengano eseguiti in maniera sequenziale e che accedano in maniera mutualmente esclusiva ad n, quindi l'unica variazione possibile è il thread di inizio per la sequenza.
Nel caso sia A ad accedere per primo alla CS si ha: n = ((1*2)+2)*3 = 12
Nel caso sia invece B ad accedere per primo alla C si ha: n = (((0*2)+1)*3)+2 = 5
Quindi n = 12 e n = 5 sono i due possibili valori.