The dinosaur book says that a solution to critical section problem must satisfy Mutual exclusion, Progress and Bounded Wait
This is the structure of a process as described under Peterson's solution in the book:
do {
flag[i]=True;
turn=j;
while (flag[j] && turn==j);
// critical section
flag[i]=False;
// remainder section
} while (True);
I dont understand how this is solving bounded waiting problem. The bounded waiting says that there is a limit to how many times a process can be stopped from getting into its critical section so that no process gets starved. But here there is no counter for that and processes share just these two variables among themselves in this solution:
int turn;
boolean flag[2];
Bounded waiting says that a bound must exist on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted.
Here, the Peterson's solution is considers strict alternation so, alternatively process[0] and process[1] will get access to critical section.Here bounded waiting won't be satisfied in case e.g. some process gets C.S. repeatedly starving other processes but this situation is not possible because of strict alternation.
By using 'turn' variable bounded waiting is ensured.
First of all it is needed to know that the Peterson's solution is a 2 process solution.
Now the answer...
Here you can see that when the process enters the loop
while(flag[j] && turn==j);
it lets the process j to enters its critical section. Here the process i will only enter its critical section when either the turn != j or flag[j] == false;
Lets say that flag[j] = true. In this case process i has to wait and can't enter its critical section(process i is waiting). Now we know that as soon as the process j is done with its critical section it executes the line
flag[j] = false;
which helps process i to get out of the loop and even if now process j again tries to enter its critical section, it will get stuck in the same loop and process i will be able to execute its critical section without waiting any longer(here the bound on waiting is 1).
Here we can see that even if process j is fast and tries to enter its critical section as many times it wants, process i wont get starved once it is ready to execute its critical section. Thus bounded waiting i.e. there is a bound on the amount of process(which in here is 1) that can execute its critical section before the process requested for, is granted
permission to execute its critical section.
Related
Consider the following code
//proces i: //proces j:
flag[i] = true; flag[j] = true;
turn = j; turn = i;
while(flag[j] == true && turn==j); while(flag[i] == true && turn == i);
<critical section> <critical section>
flag[i] = false; flag[j] = false;
<remainder section <remainder section>
I am certain that the above code will satisfies the mutual exclusion property but what I am uncertain about the following
What exactly does progress mean ? and does the above code satisfy it, The above code requires the the critical section being executed in strict alternation. Is that considered as progress ?
From what I see the above code does not maintain any information on the number of times a process has entered the critical section, would that mean that the above code does not satisfy bounded waiting ?
Progress means that the process will eventually do some work - an example of where this may not be the case is when a low-priority thread might be pre-empted and rolled back by high-priority threads. Once your processes reach their critical section they won't be pre-empted, so they'll make progress.
Bounded waiting means that the process will eventually gain control of the processor - an example of where this may not be the case is when another process has a non-terminating loop in a critical section with no possibility of the thread being interrupted. Your code has bounded waiting IF the critical sections terminate AND the remainder section will not re-invoke the process's critical section (otherwise a process might keep running its critical section without the other process ever gaining control of the processor).
Progress of the processes mean that the processes don't enter in a deadlock situation and hence,their execution continues independently! Actually, at any moment of time,only one of the process i or process j will be executing its critical section code and hence,the consistency will be maintained! SO, Progress of both processes are being talked and met successfully in the given code.
Next, this particular code is for processes which are intended to run only for once and hence, they won't be reaching their critical section code again. It is for single execution of process.
Bounded waiting says that a bound must exist on the number of times
that other processes are allowed to enter their critical sections
after a process has made a request to enter its critical section and
before that request is granted.
This particular piece of code has nothing to do with bounded waiting and is for trivial cases where processes execute for once only!
I have the following code:
pthread_mutex lock_row[M], lock_culm[M];
FUNCTION SIGNATURE (..., int i, int j, ...) {
pthread_mutex_lock(&lock_row[i]);
pthread_mutex_lock(&lock_culm[j]);
...CRITICAL CODE...
pthread_mute_unlock(&lock_row[j]);
pthread_mute_unlock(&lock_row[i]);
}
Can I get a deadlock between the first lock to the second? Let's say if we have a context switch after the first row, and other thread tries to lock something again? I don't really get this I would like to understand this a little further.
Besides the probable typo when you try to unlock sth twice, this example will never deadlock. Context switches between the two lock-calls pose no threat to the mechanism involved here. Think of it as a getting a higher level of allowance. With each lock gained, this process or thread is allowed to do more. Each locking is a gate which might hold the process up until no other lock-holder prevents the entering of the higher level. Whatever happens between the two lockings does not matter as long as it does not change that level of allowance.
pthread_mutex_lock(&lock_row[i]);
pthread_mutex_lock(&lock_culm[j]);
This is fine as long as all of your code takes these locks in this order - the lock_row lock first, then the lock_culm lock second. If another part of the code takes these same locks in the opposite order, then it can deadlock.
For this reason it is usual in complex programs to define the locking order - a global ordering of all the locks in the program, defining the order in which they should be taken.
I'm looking into the Reusable Barrier algorithm from the book "The Little Book Of Semaphores" (archived here).
The puzzle is on page 31 (Basic Synchronization Patterns/Reusable Barrier), and I have come up with a 'solution' (or not) which differs from the solution from the book (a two-phase barrier).
This is my 'code' for each thread:
# n = 4; threads running
# semaphore = n max., initialized to 0
# mutex, unowned.
start:
mutex.wait()
counter = counter + 1
if counter = n:
semaphore.signal(4) # add 4 at once
counter = 0
mutex.release()
semaphore.wait()
# critical section
semaphore.release()
goto start
This does seem to work, I've even inserted different sleep timers into different sections of the threads, and they still wait for all the threads to come before continuing each and every loop. Am I missing something? Is there a condition that this will fail?
I've implemented this using the Windows library Semaphore and Mutex functions.
Update:
Thank you to starblue for the answer. Turns out that if for whatever reason a thread is slow between mutex.release() and semaphore.wait() any of the threads that arrive to semaphore.wait() after a full loop will be able to go through again, since there will be one of the N unused signals left.
And having put a Sleep command for thread number 3, I got this result where one can see that thread 3 missed a turn the first time, with thread 1 having done 2 turns, and then catching up on the second turn (which was in fact its 1st turn).
Thanks again to everyone for the input.
One thread could run several times through the barrier while some other thread doesn't run at all.
What is the difference between a dead lock and a race around condition in programming terms?
Think of a race condition using the traditional example. Say you and a friend have an ATM cards for the same bank account. Now suppose the account has $100 in it. Consider what happens when you attempt to withdraw $10 and your friend attempts to withdraw $50 at exactly the same time.
Think about what has to happen. The ATM machine must take your input, read what is currently in your account, and then modify the amount. Note, that in programming terms, an assignment statement is a multi-step process.
So, label both of your transactions T1 (you withdraw $10), and T2 (your friend withdraws $50). Now, the numbers below, to the left, represent time steps.
T1 T2
---------------- ------------------------
1. Read Acct ($100)
2. Read Acct ($100)
3. Write New Amt ($90)
4. Write New Amt ($50)
5. End
6. End
After both transactions complete, using this timeline, which is possible if you don't use any sort of locking mechanism, the account has $50 in it. This is $10 more than it should (your transaction is lost forever, but you still have the money).
This is a called race condition. What you want is for the transaction to be serializable, that is in no matter how you interleave the individual instruction executions, the end result will be the exact same as some serial schedule (meaning you run them one after the other with no interleaving) of the same transactions. The solution, again, is to introduce locking; however incorrect locking can lead to dead lock.
Deadlock occurs when there is a conflict of a shared resource. It's sort of like a Catch-22.
T1 T2
------- --------
1. Lock(x)
2. Lock(y)
3. Write x=1
4. Write y=19
5. Lock(y)
6. Write y=x+1
7. Lock(x)
8. Write x=y+2
9. Unlock(x)
10. Unlock(x)
11. Unlock(y)
12. Unlock(y)
You can see that a deadlock occurs at time 7 because T2 tries to acquire a lock on x but T1 already holds the lock on x but it is waiting on a lock for y, which T2 holds.
This bad. You can turn this diagram into a dependency graph and you will see that there is a cycle. The problem here is that x and y are resources that may be modified together.
One way to prevent this sort of deadlock problem with multiple lock objects (resources) is to introduce an ordering. You see, in the previous example, T1 locked x and then y but T2 locked y and then x. If both transactions adhered here to some ordering rule that says "x shall always be locked before y" then this problem will not occur. (You can change the previous example with this rule in mind and see no deadlock occurs).
These are trivial examples and really I've just used the examples you may have already seen if you have taken any kind of undergrad course on this. In reality, solving deadlock problems can be much harder than this because you tend to have more than a couple resources and a couple transactions interacting.
As always, use Wikipedia as a starting point for CS concepts:
http://en.wikipedia.org/wiki/Deadlock
http://en.wikipedia.org/wiki/Race_condition
A deadlock is when two (or more) threads are blocking each other. Usually this has something to do with threads trying to acquire shared resources. For example if threads T1 and T2 need to acquire both resources A and B in order to do their work. If T1 acquires resource A, then T2 acquires resource B, T1 could then be waiting for resource B while T2 was waiting for resource A. In this case, both threads will wait indefinitely for the resource held by the other thread. These threads are said to be deadlocked.
Race conditions occur when two threads interact in a negatve (buggy) way depending on the exact order that their different instructions are executed. If one thread sets a global variable, for example, then a second thread reads and modifies that global variable, and the first thread reads the variable, the first thread may experience a bug because the variable has changed unexpectedly.
Deadlock :
This happens when 2 or more threads are waiting on each other to release the resource for infinite amount of time.
In this the threads are in blocked state and not executing.
Race/Race Condition:
This happens when 2 or more threads run in parallel but end up giving a result which is wrong and not equivalent if all the operations are done in sequential order.
Here all the threads run and execute there operations.
In Coding we need to avoid both race and deadlock condition.
I assume you mean "race conditions" and not "race around conditions" (I've heard that term...)
Basically, a dead lock is a condition where thread A is waiting for resource X while holding a lock on resource Y, and thread B is waiting for resource Y while holding a lock on resource X. The threads block waiting for each other to release their locks.
The solution to this problem is (usually) to ensure that you take locks on all resources in the same order in all threads. For example, if you always lock resource X before resource Y then my example can never result in a deadlock.
A race condition is something where you're relying on a particular sequence of events happening in a certain order, but that can be messed up if another thread is running at the same time. For example, to insert a new node into a linked list, you need to modify the list head, usually something like so:
newNode->next = listHead;
listHead = newNode;
But if two threads do that at the same time, then you might have a situation where they run like so:
Thread A Thread B
newNode1->next = listHead
newNode2->next = listHead
listHead = newNode2
listHead = newNode1
If this were to happen, then Thread B's modification of the list will be lost because Thread A would have overwritten it. It can be even worse, depending on the exact situation, but that's the basics of it.
The solution to this problem is usually to ensure that you include the proper locking mechanisms (for example, take out a lock any time you want to modify the linked list so that only one thread is modifying it at a time).
Withe rest to Programming language if you are not locking shared resources and are accessed by multiple threads then its called as "Race condition", 2nd case if you locked the resources and sequences of access to shared resources are not defined properly then threads may go long waiting for the resources to use then its a case of "deadlock"
I have this POSIX thread:
void subthread(void)
{
while(!quit_thread) {
// do something
...
// don't waste cpu cycles
if(!quit_thread) usleep(500);
}
// free resources
...
// tell main thread we're done
quit_thread = FALSE;
}
Now I want to terminate subthread() from my main thread. I've tried the following:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread);
But it does not work! The while() clause does never exit although my subthread clearly sets quit_thread to FALSE after having freed its resources!
If I modify my shutdown code like this:
quit_thread = TRUE;
// wait until subthread() has cleaned its resources
while(quit_thread) usleep(10);
Then everything is working fine! Could someone explain to me why the first solution does not work and why the version with usleep(10) suddenly works? I know that this is not a pretty solution. I could use semaphores/signals for this but I'd like to learn something about multithreading, so I'd like to know why my first solution doesn't work.
Thanks!
Without a memory fence, there is no guarantee that values written in one thread will appear in another. Most of the pthread primitives introduce a barrier, as do several system calls such as usleep. Using a mutex around both the read and write introduces a barrier, and more generally prevents multi-byte values being visible in partially written state.
You also need to separate the idea of asking a thread to stop executing, and reporting that it has stopped, and appear to be using the same variable for both.
What's most likely to be happening is that your compiler is not aware that quit_thread can be changed by another thread (because C doesn't know about threads, at least at the time this question was asked). Because of that, it's optimising the while loop to an infinite loop.
In other words, it looks at this code:
quit_thread = TRUE;
while(quit_thread);
and thinks to itself, "Hah, nothing in that loop can ever change quit_thread to FALSE, so the coder obviously just meant to write while (TRUE);".
When you add the call to usleep, the compiler has another think about it and assumes that the function call may change the global, so it plays it safe and doesn't optimise it.
Normally you would mark the variable as volatile to stop the compiler from optimising it but, in this case, you should use the facilities provided by pthreads and join to the thread after setting the flag to true (and don't have the sub-thread reset it, do that in the main thread after the join if it's necessary). The reason for that is that a join is likely to be more efficient than a continuous loop waiting for a variable change since the thread doing the join will most likely not be executed until the join needs to be done.
In your spinning solution, the joining thread will most likely continue to run and suck up CPU grunt.
In other words, do something like:
Main thread Child thread
------------------- -------------------
fStop = false
start Child Initialise
Do some other stuff while not fStop:
fStop = true Do what you have to do
Finish up and exit
join to Child
Do yet more stuff
And, as an aside, you should technically protect shared variables with mutexes but this is one of the few cases where it's okay, one-way communication where half-changed values of a variable don't matter (false/not-false).
The reason you normally mutex-protect a variable is to stop one thread seeing it in a half-changed state. Let's say you have a two-byte integer for a count of some objects, and it's set to 0x00ff (255).
Let's further say that thread A tries to increment that count but it's not an atomic operation. It changes the top byte to 0x01 but, before it gets a chance to change the bottom byte to 0x00, thread B swoops in and reads it as 0x01ff.
Now that's not going to be very good if thread B want to do something with the last element counted by that value. It should be looking at 0x0100 but will instead try to look at 0x01ff, the effect of which will be wrong, if not catastrophic.
If the count variable were protected by a mutex, thread B wouldn't be looking at it until thread A had finished updating it, hence no problem would occur.
The reason that doesn't matter with one-way booleans is because any half state will also be considered as true or false so, if thread A was halfway between turning 0x0000 into 0x0001 (just the top byte), thread B would still see that as 0x0000 (false) and keep going (until thread A finishes its update next time around).
And if thread A was turning the boolean into 0xffff, the half state of 0xff00 would still be considered true by thread B so it would do its thing before thread A had finished updating the boolean.
Neither of those two possibilities is bad simply because, in both, thread A is in the process of changing the boolean and it will finish eventually. Whether thread B detects it a tiny bit earlier or a tiny bit later doesn't really matter.
The while(quite_thread); is using the value quit_thread was set to on the line before it. Calling a function (usleep) induces the compiler to reload the value on each test.
In any case, this is the wrong way to wait for a thread to complete. Use pthread_join instead.
You're "learning" multhithreading the wrong way. The right way is to learn to use mutexes and condition variables; any other solution will fail under some circumstances.