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#include<stdio.h> | |
#include<pthread.h> | |
#include<semaphore.h> | |
sem_t mutex,writeblock; | |
int data = 0,rcount = 0; | |
void *reader(void *arg) | |
{ | |
int f; | |
f = ((int)arg); | |
sem_wait(&mutex); | |
rcount = rcount + 1; | |
if(rcount==1) | |
sem_wait(&writeblock); | |
sem_post(&mutex); | |
printf("Data read by the reader%d is %d\n",f,data); | |
sleep(1); | |
sem_wait(&mutex); | |
rcount = rcount - 1; | |
if(rcount==0) | |
sem_post(&writeblock); | |
sem_post(&mutex); | |
} | |
void *writer(void *arg) | |
{ | |
int f; | |
f = ((int) arg); | |
sem_wait(&writeblock); | |
data++; | |
printf("Data writen by the writer%d is %d\n",f,data); | |
sleep(1); | |
sem_post(&writeblock); | |
} | |
int main() | |
{ | |
int i,b; | |
pthread_t rtid[5],wtid[5]; | |
sem_init(&mutex,0,1); | |
sem_init(&writeblock,0,1); | |
for(i=0;i<=2;i++) | |
{ | |
pthread_create(&wtid[i],NULL,writer,(void *)i); | |
pthread_create(&rtid[i],NULL,reader,(void *)i); | |
} | |
for(i=0;i<=2;i++) | |
{ | |
pthread_join(wtid[i],NULL); | |
pthread_join(rtid[i],NULL); | |
} | |
return 0; | |
} |
Is it reader's priority or writer's priority??
Can you please explain with the lines of code in the writer function how it is the writers priority?
Can you please explain with the lines of code in the writer function how it is the writers priority?
sem_wait(&mutex); rcount = rcount + 1; if(rcount==1) sem_wait(&writeblock);
Inside reader you can see this piece of code, observe carefully:
rcount is 0 initially,
after statement :rcount = rcount+1;
value of rcount is 1,
If rcount is 1, It is calling sem_wait(&writeblock), this will make sure that the read block is on hold for long enough time.
Since now read block is on hold, the write block will be executed.
I hope I have answered your question.
Thank You!
How can I solve this problem using shared_mutex?
it is binary semaphore
Loops are used to implement multiple reader and writer.