# Process Synchronization

Wikipedia: Thread synchronization is defined as a mechanism which ensures that two or more concurrent processes or threads do not simultaneously execute some particular program segment known as critical section.

## Race Condition & Critical Section Problem

Consider the following producer and consumer programs:

//producer process
while(true){
while(counter==BUFFER_SIZE){
// buffer is full, do nothing.
}
buffer[in] = next.produced;
in = (in+1)%BUFFER_SIZE;
counter++;
}

//consumer process
while(true){
while(counter==0){
// buffer is empty, nothing to consume. do nothing.
}
next_consumed = buffer[out];
out = (out+1)% BUFFER_SIZE;
counter--;
}


Suppose the consumer and producer processes are run concurrently, at it just so happens that producer process enters the counter++ row at about the same time as the consumer process entering the counter-- row, a problem might occur.

To execute the decrementation or incrementation of a variable, the cpu fetches the data from the memory location of the variable counter and store it in a register. It then increments (or decrements) the register by one and then save the data back into the memory location of counter variable.

What’s wrong here?

Let’s imagine that when it happens there are 4 items in the buffer and BUFFER_SIZE is equal to 10. Logically speaking, The producer would like to increment the counter by one so that counter should then be 5, and the consumer would decrement the counter after consumption so that the counter ends up being 4.

However, since the programs are run concurrently, when both programs execute the incrementation/decrementation rows, the sequence of the underlying operations might look like

T0: producer register1 = counter // register1 is 4 and counter is 4
T1: producer register1 = register + 1 // register1 is 5, counter is still 4.
T2: consumer register2 = counter // register2 is 4 and counter is still 4
T3: producer counter = register1 // counter is now 5
T4: consumer register2 = register2 - 1 // register2 is now 3
T5: consumer counter = register2 // counter is now 3

Did you see it? In the end the counter is incorrectly set to 3 instead of what it should be (4).

When multiple processes access and manipulate the same data concurrently and the correct outcome depends on particular sequence of access/manipulations, it becomes a potential bug called race condition.

The critical section problem is introduced to solve this kind of issues and it aims to design a protocol for each process to request permission before it enters its critical section, within which process would access and manipulate shared data.

A qualified solution must satisfy

1. mutual exclusion: when one process is in its critical section, no other processs should be in its critical section

2. progress: the selection of the process that enters its critical section cannot be postponed indefinitely, and therefore progress is made in each process.

3. bounded waiting: there is a bound that limits the number of times process A enters its critical section after a process B requests to enter its critical section.

## Peterson’s Solution

Two processes share int turn; and boolean flag[2];.

do{
flag[i] = true;
turn = j;
while(flag[j] && turn==j){
// do nothing, process j is in its critical section
}
//critical section, do something
flag[i] = false;
//remainder section, do something
}while(true)


Mutual Exclusion (Mutex)

For process i to be in its critical section flag[j] must be false or turn==j must be false. Since turn cannot be j and i at the same time, mutual exclusion is achieved.

Progress & Bounded Waiting

Process i can be prevented from entering its critical section if flag[j] stays true and turn==j stays true.

Imagine that now process i is stuck in while loop waiting for process j, there are several cases to be examined:

1. process j is running the 1st line of code setting flag[j]=false and turn=i, so process i can then enter its critical section.

2. process j is executing the while loop too. in this case, either turn==i or turn===j , if turn is i, then i enters its critical section; if turn is j, then j enters its critical section, finishes and set flag[j]=false, which in turn allows process i to enter its critical section.

Thus, process i will enter its critical section (progress) after at most one time of process j’s entering its critical section (bounded waiting)

## Hardware Solution

Hardware instructions that allow programmers to access/manipulate data atomically can help solve the critical section problem. By atomically, it means that the instructions are run as one uninterruptible unit.

An example of such hardware instruction:

boolean test_and_set(boolean* target){
boolean rv = *target;
*target = true;
return rv;
}


Given that test_and_set can be run atomically, it can be used to solve critical section problem like this:


boolean available = false;
do {
while(!test_and_set(available)){
//lock not available, waiting
}
// critical section
available = true;
} while(true);


Another example is compare_and_swap:

int compare_and_swap(int *value, int expected, int new_value){
int temp = *value;
if (*value == expected){
*value = new_value;
}
return temp;
}

do {
while(compare_and_swap(&lock, 0, 1) != 0){
//do nothing
}
//critical section
lock = 0;
//remainder section
} while(true)


In this example, if 2 processes are run concurrently and both run the while statement. Since the function compare_and_swap is run atomically, one process would go first and the second one follows after the first one exits compare_and_swap. At the beginning the lock is initiated as 0, so the first process that runs compare_and_swap is able to change the value of the lock ((*value==expected) is true) and return its initial value of 0, thereby allowing process 1 to exit the while loop and enter its critical section. Process 2 now enters compare_and_swap and since *value != expected it is trapped in the while loop until process 1 finishes its critical section and set the lock to 0.

## Software Solution - Mutex Lock

With acquire() and release() functions, a process has to acquire the lock before entering its critical section. If the local is not available, the process busy waits until the lock becomes available.

acquire(){
while(!available){
//do nothing, wait
}
available = false;
}

release(){
available = true;
}


Both methods have to be able to be run atomically for it to work, therefore some hardware instructions that are run atomically can be of help to implement acquire and release.

Assuming we have a struct lock that has a variable int available, we can implement the mutex lock as shown below:

typedef struct {
int available;
} lock;

void acquire(lock* mutex){
while (compare_and_swap(mutex->available, 0, 1)!=0){
//lock is not available, waiting
}
}

void release(lock *mutex){
test_and_set(mutex->available);
}


The implementation has a major drawback: spinlock. Spinlock means that when a process is in its critical section, other waiting processes are in a continuous while loop (therefore spinning busily).

The busy waiting wastes CPU cycle. However busy busy wait does not require context switch, which requires a significant amount of time. Thus if the process spins only for a short period of time, it’s actually more advantageous to busy wait.

Spinlocks are especially common for multi-processor systems because a thread can spin on one processor while another thread perform s its critical section on another processor.

## Software Solution - Semaphore

Semaphores help sync processes by incrementing and decrementing. There are two functions that work with semaphores: signal and wait

wait (S){
while (S<=0){
//no resource, waiting
}
S--;
}

signal(S){
S++;
}


For counting semaphores, their values can be in the range of all integers. Let’s say for example there are 10 resources that are shared by n process where $n \geq 10$. The semaphore S is therefore initiated to 10 and each process calls wait(S) before executing its critical section which manipulates one resource, and calls signal when it has finished using the resource.

The implementation above still suffers from spinlocks (the while loop); to improve it, we can revise the implementation as :

typedef struct {
int value;
struct process* queue;
}semaphore;

wait (S){
S--;
if (S<0){
block();
}
}

signal(S){
S++;
if (S->value<=0){
dequeue one process from S->queue
wakeup the process
}
}


With the previous implementation of semaphores, we are introduced with two new concepts: deadlocks and starvation.

For example, given two processes P1 and P2, they might be executing:

//process P1
wait(S1);
wait(S2);

signal(S1);
signal(S2);

//process P2
wait(S2);
wait(S1);

signal(S2);
signal(S1);


In this example, let’s assume S1 and S2 are binary semaphores.

If P1 and P2 are run concurrently and both successfully pass the first line of their codes. P1 will then be blocked in wait(S2) because P2 has obtained the lock of S2. P2 will then be blocked in wait(S1) because P1 has obtained the lock of S1.

Both are now trapped and can never get out because P1 relies on P2 calling signal(S2) and P2 relies on P1 calling signal(S1) to be waken up.

Situations, where every process is waiting for an event (e.g. releasing the resource access) that can only be caused by another process, are called deadlocks.

Starvation, on the other hand, happens when a process waits indefinitely in a semaphore, which might occur when the waiting list of the semaphore is implemented in a stack-like manner (last-in-first-out).

## Priority Inversion

let L, M, H be three process with priorities Lp < Mp < Hp. Suppose L is currently using resource R and H requires access to R too and is waiting for L to release the resource. At the same time, M becomes available to run and preempting L. This allows M to be able to indirectly affect how long H waits for L to release R, which is called priority inversion.

A priority inheritance protocol can be implemented to resolve this problem. With this protocol, all processing accessing resources needed by a higher priority process inherit the higher priority until they are finished with the resources.

Let’s look at the previous example again: L is using R, and H comes in and requests for access to R. L now has temporarily inherits the Hp priority, and when M becomes runnable, it cannot preempt a process with higher priority Hp that L current represents. When L finishes with R, it retains its original process priority Lp, H gets access to R and starts running while M keeps waiting until L is done.