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March 19, 2015 16:32
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Semaphore - Reader Writer Problem
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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; | |
} |
It is first writers priority. Once the writer has written something then
only any reader can read.
…On Fri, 27 Sep 2019, 1:49 pm Ferogle, ***@***.***> wrote:
Is it reader's priority or writer's priority??
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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!
I thought that would force write to wait. Like I thought it would force
write to not be capable of executing if there were new writers after and
that it would need to wait for the reader to finish.
…On Fri, Apr 3, 2020 at 7:59 AM Mritunjay Choubey ***@***.***> wrote:
***@***.**** commented on this gist.
------------------------------
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!
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How can I solve this problem using shared_mutex?
I figured it out. Thank you.
…On Fri, Apr 10, 2020 at 10:38 AM jprustom ***@***.***> wrote:
***@***.**** commented on this gist.
------------------------------
How can I solve this problem using shared_mutex?
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it is binary semaphore
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Is it reader's priority or writer's priority??