Explanation:
Below is required code in java :-
public class Table{
private String color; //attribute to store the color of the table
public Table(){ //default constructor
this.color=""; //set a default blank color
}
public Table(String color){ //overloaded constructor
this.color=color; //set the color value equal to the parameter provided
}
public void setColor(String color){ //setter or mutator method
this.color=color; //set the color value equal to the parameter provided
}
}
Answer:
170 hope this helps you with
Answer:
int sumid=0; /* Shared var that contains the sum of the process ids currently accessing the file */
int waiting=0; /* Number of process waiting on the semaphore OkToAccess */
semaphore mutex=1; /* Our good old Semaphore variable ;) */
semaphore OKToAccess=0; /* The synchronization semaphore */
get_access(int id)
{
sem_wait(mutex);
while(sumid+id > n) {
waiting++;
sem_signal(mutex);
sem_wait(OKToAccess);
sem_wait(mutex);
}
sumid += id;
sem_signal(mutex);
}
release_access(int id)
{
int i;
sem_wait(mutex);
sumid -= id;
for (i=0; i < waiting;++i) {
sem_signal(OKToAccess);
}
waiting = 0;
sem_signal(mutex);
}
main()
{
get_access(id);
do_stuff();
release_access(id);
}
Some points to note about the solution:
release_access wakes up all waiting processes. It is NOT ok to wake up the first waiting process in this problem --- that process may have too large a value of id. Also, more than one process may be eligible to go in (if those processes have small ids) when one process releases access.
woken up processes try to see if they can get in. If not, they go back to sleep.
waiting is set to 0 in release_access after the for loop. That avoids unnecessary signals from subsequent release_accesses. Those signals would not make the solution wrong, just less efficient as processes will be woken up unnecessarily.
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Answer:
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