Question

1. Monitor Exercise: A file is to be shared among different processes, each of which has a unique priority number. The file can be accessed simultaneously by several processes, subject to the following constraints: The maximum number of processes accessing the file is 20 and the priority number of all the processes currently accessing the files must be greater than the priority number of the process requesting access. Add comments to describe your pseudocode solution. Using pseudocode write a monitor to coordinate access to the file. (6 points) See the Module 9 - Notes part 3 - Monitor Example.pdf AND Module 9-Notes part 3-Monitor Example2.pdf A basic monitor is set up as follows (see examples on Canvas) Monitor name f //shared variables REQUESTprocedure RELEASEprocedure ) //initializtion code)(

Answer 1

Program:

#include <stdio.h>
int n=10;// file id number
int sumOfID=0; / Shared var that contains the sum of the process ids currently accessing the file /
int numOfProcessWaiting=0; / Number of process waiting on the semaphore okToAccess /
semaphore mutex=1; / Semaphore variable ;) /
semaphore okToAccess=0; / The synchronization semaphore /

void get_access(int pid)
{
sem_wait(mutex);
/*all unique numbers associated with all the processes currently accessing the file must be less than n*/
while(sumOfID+pid > n)
{
numOfProcessWaiting++;
sem_signal(mutex);
sem_wait(okToAccess);
sem_wait(mutex);
}
sumOfID += pid;
sem_signal(mutex);
}
void release_access(int pid)
{
int i;
sem_wait(mutex);
sumOfID -= pid;
for (i=0; i < numOfProcessWaiting;++i) {
sem_signal(okToAccess);
}
numOfProcessWaiting = 0;
sem_signal(mutex);
}

void Monitor()
{
int id;
for(id=0;id<20;id++)
{
get_access(id);
}

for(id=0;id<20;id++)
{
release_access(id);
}
}
int main()
{
Monitor();
}

Note: Use "Semaphore class" code with this file to get execution sucess.

As per your requirement following above code is written along with comments for each and important line of codes.