- 读者-写者问题
允许多个进程同时对数据进行读操作，但是不允许读和写以及写和写操作同时发生。
一个整型变量 count 记录在对数据进行读操作的进程数量，一个互斥量 count_mutex 用于对 count 加锁，一个互斥量 data_mutex 用于对读写的数据加锁。
typedef int semaphore;
semaphore count_mutex = 1;
semaphore data_mutex = 1;
int count = 0;
void reader() {
while(TRUE) {
down(&count_mutex);
count++;
if(count == 1) down(&data_mutex); // 第一个读者需要对数据进行加锁，防止写进程访问
up(&count_mutex);
read();
down(&count_mutex);
count--;
if(count == 0) up(&data_mutex);
up(&count_mutex);
}
}
void writer() {
while(TRUE) {
down(&data_mutex);
write();
up(&data_mutex);
}
}
以下内容由 @Bandi Yugandhar 提供。
The first case may result Writer to starve. This case favous Writers i.e no writer, once added to the queue, shall be kept waiting longer than absolutely necessary(only when there are readers that entered the queue before the writer).
int readcount, writecount; //(initial value = 0)
semaphore rmutex, wmutex, readLock, resource; //(initial value = 1)
//READER
void reader() {
<ENTRY Section>
down(&readLock); // reader is trying to enter
down(&rmutex); // lock to increase readcount
readcount++;
if (readcount == 1)
down(&resource); //if you are the first reader then lock the resource
up(&rmutex); //release for other readers
up(&readLock); //Done with trying to access the resource
<CRITICAL Section>
//reading is performed
<EXIT Section>
down(&rmutex); //reserve exit section - avoids race condition with readers
readcount--; //indicate you're leaving
if (readcount == 0) //checks if you are last reader leaving
up(&resource); //if last, you must release the locked resource
up(&rmutex); //release exit section for other readers
}
//WRITER
void writer() {
<ENTRY Section>
down(&wmutex); //reserve entry section for writers - avoids race conditions
writecount++; //report yourself as a writer entering
if (writecount == 1) //checks if you're first writer
down(&readLock); //if you're first, then you must lock the readers out. Prevent them from trying to enter CS
up(&wmutex); //release entry section
<CRITICAL Section>
down(&resource); //reserve the resource for yourself - prevents other writers from simultaneously editing the shared resource
//writing is performed
up(&resource); //release file
<EXIT Section>
down(&wmutex); //reserve exit section
writecount--; //indicate you're leaving
if (writecount == 0) //checks if you're the last writer
up(&readLock); //if you're last writer, you must unlock the readers. Allows them to try enter CS for reading
up(&wmutex); //release exit section
}
We can observe that every reader is forced to acquire ReadLock. On the otherhand, writers doesn’t need to lock individually. Once the first writer locks the ReadLock, it will be released only when there is no writer left in the queue.
From the both cases we observed that either reader or writer has to starve. Below solutionadds the constraint that no thread shall be allowed to starve; that is, the operation of obtaining a lock on the shared data will always terminate in a bounded amount of time.
int readCount; // init to 0; number of readers currently accessing resource
// all semaphores initialised to 1
Semaphore resourceAccess; // controls access (read/write) to the resource
Semaphore readCountAccess; // for syncing changes to shared variable readCount
Semaphore serviceQueue; // FAIRNESS: preserves ordering of requests (signaling must be FIFO)
void writer()
{
down(&serviceQueue); // wait in line to be servicexs
// <ENTER>
down(&resourceAccess); // request exclusive access to resource
// </ENTER>
up(&serviceQueue); // let next in line be serviced
// <WRITE>
writeResource(); // writing is performed
// </WRITE>
// <EXIT>
up(&resourceAccess); // release resource access for next reader/writer
// </EXIT>
}
void reader()
{
down(&serviceQueue); // wait in line to be serviced
down(&readCountAccess); // request exclusive access to readCount
// <ENTER>
if (readCount == 0) // if there are no readers already reading:
down(&resourceAccess); // request resource access for readers (writers blocked)
readCount++; // update count of active readers
// </ENTER>
up(&serviceQueue); // let next in line be serviced
up(&readCountAccess); // release access to readCount
// <READ>
readResource(); // reading is performed
// </READ>
down(&readCountAccess); // request exclusive access to readCount
// <EXIT>
readCount--; // update count of active readers
if (readCount == 0) // if there are no readers left:
up(&resourceAccess); // release resource access for all
// </EXIT>
up(&readCountAccess); // release access to readCount
}
- 哲学家进餐问题
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/a9077f06-7584-4f2b-8c20-3a8e46928820.jpg"/> </div><br>
五个哲学家围着一张圆桌，每个哲学家面前放着食物。哲学家的生活有两种交替活动：吃饭以及思考。当一个哲学家吃饭时，需要先拿起自己左右两边的两根筷子，并且一次只能拿起一根筷子。
下面是一种错误的解法，考虑到如果所有哲学家同时拿起左手边的筷子，那么就无法拿起右手边的筷子，造成死锁。
#define N 5
void philosopher(int i) {
while(TRUE) {
think();
take(i); // 拿起左边的筷子
take((i+1)%N); // 拿起右边的筷子
eat();
put(i);
put((i+1)%N);
}
}
为了防止死锁的发生，可以设置两个条件：
- 必须同时拿起左右两根筷子；
- 只有在两个邻居都没有进餐的情况下才允许进餐。
#define N 5
#define LEFT (i + N - 1) % N // 左邻居
#define RIGHT (i + 1) % N // 右邻居
#define THINKING 0
#define HUNGRY 1
#define EATING 2
typedef int semaphore;
int state[N]; // 跟踪每个哲学家的状态
semaphore mutex = 1; // 临界区的互斥
semaphore s[N]; // 每个哲学家一个信号量
void philosopher(int i) {
while(TRUE) {
think();
take_two(i);
eat();
put_two(i);
}
}
void take_two(int i) {
down(&mutex);
state[i] = HUNGRY;
test(i);
up(&mutex);
down(&s[i]);
}
void put_two(i) {
down(&mutex);
state[i] = THINKING;
test(LEFT);
test(RIGHT);
up(&mutex);
}
void test(i) { // 尝试拿起两把筷子
if(state[i] == HUNGRY && state[LEFT] != EATING && state[RIGHT] !=EATING) {
state[i] = EATING;
up(&s[i]);
}
}