Bounded Waiting in Test and Set Instruction - operating-system

Question

In order to gurantee Bounded wait in Test and set Instruction,following is the code given in Operating system book,Galvin -:
do {
1 waiting[i] = true;
2 while (waiting[i] && test_and_set(&lock)) ;
3 waiting[i] = false;
/* critical section */
4 j = (i + 1) % n;
5 while ((j != i) && !waiting[j])
6 j = (j + 1) % n;
7 if (j == i)
8 lock = false;
9 else
10 waiting[j] = false;
/* remainder section */
} while (true);
I am getting the complete code and concluded that
A process P_i will be in the critical section if either
Waiting [i]=false or test_and_set(&lock)=FALSE which ensures that Lock was FALSE previously. so Exit Section is either setting Waiting[j] or lock to FALSE.
But i have got some doubts-:
if in the exit section section it is found that same process again requests for critical section i.e
if j==i
then according to the code,that process have to start its execution form line number 2,i.e will execute
test_and_set(&lock))
in while loop and find the return value of test_and_set(&lock)) as false and then move to critical section.My doubt is that if same process wants to be in critical section ,is it necessary to start its exection right from line number 2
2.Now i want to do following Permutation and want to check the possible outcome.i want to swap line number 8 and 10
in line number 8 if i make
waiting[j]=false;
then also it will move to critical section even though lock =true now.
in line number 10 if i make
lock=false
then also it(process p_j) will move to critical section even though waiting[i]=true and i think it would be better because line number 3 will assign waiting[i]=false ,after the while loop breaks due to test_and_set(&lock)=false.
On the other hand if i make this change process have to execute test_and_set(&lock) which is time consuming
Is my assumption for point2 right?
what is the correct reason for point 1?
Thanks

Answer 1

Regarding point 1:
My doubt is that if same process wants to be in critical section ,is
it necessary to start its exection right from line number 2
A process is basically an program in execution. A process just cannot jump around choosing which instruction to execute next. The control flow decides that. The control of the code(which is the process itself) suggests that if a process successfully enters critical section, and again wants to enter critical section, then it will first execute lines 4 to 10 and then execute remainder section and would have to start execution right from line 1
Regarding your point 2
Now i want to do following Permutation and want to check the possible
outcome.i want to swap line number 8 and 10
If you swap the lines, the bounded waiting condition would no longer exist.
Proof
Suppose that only 1 process P(i) made the request to access Critical Section and it successfully entered. So
lock = true and waiting[i] = true
because only then it would have been able to come out of the for loop. Now it starts executing from line 4
Then j takes following values:
i + 2 , i + 3, ......0 , 1 , 2 , 3 , 4....i
wrap around of values occurs because of % operator. And Because no other process made request to enter critical section, waiting[j] = false for j != i
Therefore the condition while ((j != i) && !waiting[j]) becomes false when j equals i, and now we are at line 7. The new code is:
if (j == i)
waiting[j] = false;
else
lock = false;
Now if any process makes a request to enter critical section, then while (waiting[i] && test_and_set(&lock)) ; would always evaluate to true because lock is true and waiting[i] is also true and it would be stuck in spinlock. There would be no progress.