Readers writer problem is another example of a classic synchronization problem. There are many variants of this problem, one of which is examined below.
There is a shared resource which should be accessed by multiple processes. There are two types of processes in this context. They are reader and writer. Any number of readers can read from the shared resource simultaneously, but only one writer can write to the shared resource. When a writer is writing data to the resource, no other process can access the resource. A writer cannot write to the resource if there are non zero number of readers accessing the resource at that time.
From the above problem statement, it is evident that readers have higher priority than writer. If a writer wants to write to the resource, it must wait until there are no readers currently accessing that resource.
Here, we use one mutex m
and a semaphore w
. An integer variable read_count
is used to maintain the number of readers currently accessing the resource. The variable read_count
is initialized to 0
. A value of 1
is given initially to m
and w
.
Instead of having the process to acquire lock on the shared resource, we use the mutex m
to make the process to acquire and release lock whenever it is updating the read_count
variable.
The code for the writer process looks like this:
while(TRUE)
{
wait(w);
/* perform the write operation */
signal(w);
}
And, the code for the reader process looks like this:
while(TRUE)
{
//acquire lock
wait(m);
read_count++;
if(read_count == 1)
wait(w);
//release lock
signal(m);
/* perform the reading operation */
// acquire lock
wait(m);
read_count--;
if(read_count == 0)
signal(w);
// release lock
signal(m);
}