Multithreaded Fibonacci Pair Program - c

Question

I'm trying to write a program which creates two threads: a "front-end" and "back-end" thread. I want to create a "back-end" thread to iterate and compute pairs of terms from the fibonacci sequence and put them in an array, and a "front-end" thread that will print out the pairs of the array at each iteration.
"Front-End" Thread - For displaying result of "Back-End" thread operations in each iterations
"Back-End" Thread - For calculating and setting an array
ie. [5, 8], and after an iteration it will contain [13, 21]
I'm struggling to implement the Fibonacci sequence part in a thread and I've made the following progress:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <errno.h>
int fib;
void *front_end(void *ptr);
void *back_end(void *ptr);
int main() {
pthread_t thread1, thread2;
int arr[2] = {5,8};
const int *ptrtoarr;
ptrtoarr=arr;
int create1, create2;
int *s=(int *)(ptrtoarr);
printf("%d \n", *s);
ptrtoarr++;
s = (int *)(ptrtoarr);
printf("%d \n", *s);
ptrtoarr--;
create1 = pthread_create(&thread1, NULL, back_end, &arr);
if(create1) {
fprintf(stderr,"Error - pthread_create() return code: %d\n",create1);
exit(EXIT_FAILURE);
}
pthread_join(thread1, NULL);
//pthread_join(thread2, NULL);
}
// front-end thread to be callback for each back-end iteration
void *front_end(void *ptr) {
int *sum = ptr;
int i, upper = atoi(ptr);
if (upper > 0) {
for (i=0; i<upper; i++){
//Print the fib pairs
}
}
pthread_exit(0);
}
void *back_end(void *ptr) {
int i, upper = atoi(ptr);
fib=1;
if(upper > 0) {
int pre1 = 0;
int current;
//calc fib numbers.....
if(fib == 1){
printf("")
}
}
}
Can someone guide me through how I might approach this?

Answer 1

Your skeleton needs work.
Assuming the following:
unsigned n = ...; // How many to generate.
unsigned n_ready = 2; // How many are ready to print.
unsigned *fibs = malloc(sizeof(unsigned)*n);
fibs[0] = 0;
fibs[1] = 1;
At the core of your back end worker, you will have
for (unsigned i=2; i<n; ++i) {
fibs[i] = fibs[i-2] + fibs[i-1];
n_ready = i+1;
}
At the core of your frontend worker, you will have
for (unsigned i=0; i<n; ++i) {
while (i >= n_ready)
/* Nothing */;
printf("%u\n", fibs[i]);
}
Problem #1
You get into problems if a thread tries to read a variable when another is writing to it. Two or more threads reading the same variable at the same time is ok.
The variables used by both threads are n, the elements of fib[] and n_ready.
n:Not changed by either thread, so we don't need to control access to it.
fib[i] for i >= n_ready:Only accessed by the back end worker, so we don't need to control access to these.
fib[i] for i < n_ready:Only accessed by the frontend worker, so we don't need to control access to these.
n_ready:The back end worker could set n_ready at any time, and the frontend work could try to read n_ready at any time, so we do need to control access to n_ready.
Mutex are usually used to ensure that only one thread is accessing a resource (e.g. a variable, group of variables, file handle, etc) at a time.
Our back end worker becomes
for (unsigned i=2; i<n; ++i) {
// The mutex only protects n_ready
// --nothing else is going to touch fib[i-2] or fib[i-1] or fib[i]--
// so we don't need to obtain a lock yet.
fibs[i] = fibs[i-2] + fibs[i-1];
// We need to access n_ready.
pthread_mutex_lock(&mutex);
n_ready = i+1;
pthread_mutex_unlock(&mutex);
}
Our frontend worker becomes
for (unsigned i=0; i<n; ++i) {
// We need to access n_ready.
pthread_mutex_lock(&mutex);
while (i >= n_ready) {
// Allow other thread to gain the lock.
pthread_mutex_unlock(&mutex);
// We need to access n_ready.
pthread_mutex_lock(&mutex);
}
// The mutex only protects n_ready
// --nothing is going to change fib[i]--
// so we can release it now rather than later.
pthread_mutex_unlock(&mutex);
printf("%u\n", fibs[i]);
}
Problem #2
You have a busy loop. In general, this is bad because it means your thread is using 100% doing nothing by waiting. (In this particular case, since i >= n_ready is probably already true, this would actually be a good strategy. But let's ignore that.) A thread can sleep until signaled by another thread using condition vars.
Our back end worker becomes
for (unsigned i=2; i<n; ++i) {
// The mutex only protects n_ready
// --nothing else is going to touch fib[i-2] or fib[i-1] or fib[i]--
// so we don't need to obtain a lock yet.
fibs[i] = fibs[i-2] + fibs[i-1];
// We need to access n_ready.
pthread_mutex_lock(&mutex);
n_ready = i+1;
// Wake up the other thread if it's blocked.
pthread_cond_signal(&cond);
pthread_mutex_unlock(&mutex);
}
Our frontend worker becomes
for (unsigned i=0; i<n; ++i) {
// We need to access n_ready.
pthread_mutex_lock(&mutex);
while (i >= n_ready)
pthread_cond_wait(&cond, &mutex);
// The mutex only protects n_ready
// --nothing is going to change fib[i]--
// so we can release it now rather than later.
pthread_mutex_unlock(&mutex);
printf("%u\n", fibs[i]);
}
Always call pthread_cond_wait on a locked mutex. It will unlock the mutex when it's called, and it will lock it before returning. This allows the other thread to obtain the mutex in order to change n_ready.
Complete code:
#include <errno.h>
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#define UNUSED(x) (void)(x)
// To control access to n_ready.
static pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
static pthread_cond_t cond = PTHREAD_COND_INITIALIZER;
static unsigned n_ready = 0; // How many are ready to print.
static unsigned n; // How many to generate.
static unsigned *fibs = NULL;
static void *back_worker(void *unused) {
UNUSED(unused);
fibs[0] = 0;
fibs[1] = 1;
// We need to access n_ready.
pthread_mutex_lock(&mutex);
n_ready = 2;
// Wake up the other thread if it's blocked.
pthread_cond_signal(&cond);
pthread_mutex_unlock(&mutex);
for (unsigned i=2; i<n; ++i) {
// The mutex only protects n_ready
// --nothing is going to touch fib[i]--
// so we don't need to obtain a lock yet.
fibs[i] = fibs[i-2] + fibs[i-1];
// We need to access n_ready.
pthread_mutex_lock(&mutex);
n_ready = i+1;
// Wake up the other thread if it's blocked.
pthread_cond_signal(&cond);
pthread_mutex_unlock(&mutex);
}
return NULL;
}
static void *front_worker(void *unused) {
UNUSED(unused);
for (unsigned i=0; i<n; ++i) {
// We need to access n_ready.
pthread_mutex_lock(&mutex);
while (i >= n_ready)
pthread_cond_wait(&cond, &mutex);
// The mutex only protects n_ready
// --nothing is going to change fib[i]--
// so we can release it now rather than later.
pthread_mutex_unlock(&mutex);
printf("%u\n", fibs[i]);
}
return NULL;
}
int main(void) {
n = 20; // How many to generate.
fibs = malloc(sizeof(unsigned) * n);
pthread_t back_thread;
if (errno = pthread_create(&back_thread, NULL, back_worker, NULL)) {
perror(NULL);
exit(1);
}
pthread_t front_thread;
if (errno = pthread_create(&front_thread, NULL, front_worker, NULL)) {
perror(NULL);
exit(1);
}
pthread_join(back_thread, NULL);
pthread_join(front_thread, NULL);
pthread_cond_destroy(&cond);
pthread_mutex_destroy(&mutex);
free(fibs);
return 0;
}
Output:
$ gcc -Wall -Wextra -pedantic a.c -o a -lpthread && a
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
987
1597
2584
4181
Suggestion for an exercise to apply the above
Create a pool of workers that print out the numbers placed into a queue. The output doesn't need to be in order.
The worker function is already written for you. You may not change the main or worker functions. I've even created the queue for you. You simply have to make it thread safe by modifying Queue_enqueue, Queue_dequeue and Queue_done functions. These are the only functions you may change.
#include <errno.h>
#include <inttypes.h>
#include <pthread.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define NUM_WORKERS 4
#define QUEUE_SIZE 10
#define NUM_ITEMS 40
typedef struct {
pthread_mutex_t mutex;
pthread_cond_t cond;
int done;
int empty;
int full;
size_t max;
size_t next_insert;
size_t next_read;
unsigned *buf;
} Queue;
static void Queue_init(Queue* q, size_t max) {
pthread_mutex_init(&(q->mutex), NULL);
pthread_cond_init(&(q->cond), NULL);
q->done = 0;
q->empty = 1;
q->full = 0;
q->max = max;
q->next_insert = 0;
q->next_read = 0;
q->buf = malloc(sizeof(unsigned)*max);
}
static void Queue_destroy(Queue *q) {
free(q->buf);
pthread_cond_destroy(&(q->cond));
pthread_mutex_destroy(&(q->mutex));
}
static void Queue_done(Queue *q) {
q->done = 1;
}
// Returns the oldest item from the queue (via a parameter) and returns 1.
// If the queue is empty and done, returns 0.
// If the queue is empty and not done, waits until that changes.
static int Queue_dequeue(Queue *q, unsigned *i) {
while (q->empty && !q->done) {
}
if (q->empty) {
// We are completely done.
return 0;
} else {
*i = q->buf[ q->next_read ];
q->next_read = ( q->next_read + 1 ) % q->max;
q->empty = q->next_read == q->next_insert;
q->full = 0;
return 1;
}
}
// Adds the argument to the queue.
// If the queue is full, waits until that changes.
static void Queue_enqueue(Queue *q, unsigned i) {
while (q->full && !q->done) {
}
if (q->done) {
fprintf(stderr, "Error: Attempted to add item to \"done\" queue.\n");
return;
}
q->buf[q->next_insert] = i;
q->next_insert = ( q->next_insert + 1 ) % q->max;
q->empty = 0;
q->full = q->next_insert == q->next_read;
}
static int msleep(long msec) {
struct timespec ts;
int res;
if (msec < 0) {
errno = EINVAL;
return -1;
}
ts.tv_sec = msec / 1000;
ts.tv_nsec = (msec % 1000) * 1000000;
do {
res = nanosleep(&ts, &ts);
} while (res && errno == EINTR);
return res;
}
// Protects access to stdout.
static pthread_mutex_t stdout_mutex;
static Queue q;
static void *worker(void *worker_id_) {
uintptr_t worker_id = (uintptr_t)worker_id_;
unsigned int seed = worker_id; // Whatever.
unsigned i;
while (Queue_dequeue(&q, &i)) {
pthread_mutex_lock(&stdout_mutex);
printf("[%" PRIuPTR "] Dequeued %u\n", worker_id, i);
pthread_mutex_unlock(&stdout_mutex);
// msleep( rand_r(&seed) % 1000 + 1000 ); // Simulate a 1 to 2s load.
pthread_mutex_lock(&stdout_mutex);
printf("[%" PRIuPTR "] Finished processing %u\n", worker_id, i);
pthread_mutex_unlock(&stdout_mutex);
}
return NULL;
}
int main(void) {
Queue_init(&q, QUEUE_SIZE);
pthread_t workers[NUM_WORKERS];
for (uintptr_t i=0; i<NUM_WORKERS; ++i) {
if (errno = pthread_create(&(workers[i]), NULL, worker, (void*)i)) {
perror(NULL);
exit(1);
}
}
for (unsigned i=0; i<NUM_ITEMS; ++i) {
pthread_mutex_lock(&stdout_mutex);
printf("[x] Enqueuing %u...\n", i);
pthread_mutex_unlock(&stdout_mutex);
Queue_enqueue(&q, i);
pthread_mutex_lock(&stdout_mutex);
printf("[x] Enqueued %u.\n", i);
pthread_mutex_unlock(&stdout_mutex);
}
Queue_done(&q);
pthread_mutex_lock(&stdout_mutex);
printf("[x] Called done.\n");
pthread_mutex_unlock(&stdout_mutex);
for (unsigned i=0; i<NUM_WORKERS; ++i)
pthread_join(workers[i], NULL);
Queue_destroy(&q);
pthread_mutex_destroy(&stdout_mutex);
return 0;
}
If you have questions about this, feel free to post a link to the question as a comment to this answer.
Solution to suggested excercise:
static void Queue_done(Queue *q) {
pthread_mutex_lock(&(q->mutex));
q->done = 1;
pthread_cond_signal(&(q->cond));
pthread_mutex_unlock(&(q->mutex));
}
// Returns the oldest item from the queue (via a parameter) and returns 1.
// If the queue is empty and done, returns 0.
// If the queue is empty and not done, waits until that changes.
static int Queue_dequeue(Queue *q, unsigned *i) {
pthread_mutex_lock(&(q->mutex));
while (q->empty && !q->done)
pthread_cond_wait(&(q->cond), &(q->mutex));
int dequeued;
if (q->empty) {
// We are completely done.
dequeued = 0;
} else {
*i = q->buf[ q->next_read ];
q->next_read = ( q->next_read + 1 ) % q->max;
q->empty = q->next_read == q->next_insert;
q->full = 0;
dequeued = 1;
}
pthread_cond_signal(&(q->cond));
pthread_mutex_unlock(&(q->mutex));
return dequeued;
}
// Adds the argument to the queue.
// If the queue is full, waits until that changes.
static void Queue_enqueue(Queue *q, unsigned i) {
pthread_mutex_lock(&(q->mutex));
while (q->full && !q->done)
pthread_cond_wait(&(q->cond), &(q->mutex));
if (q->done) {
fprintf(stderr, "Error: Attempted to add item to \"done\" queue.\n");
} else {
q->buf[q->next_insert] = i;
q->next_insert = ( q->next_insert + 1 ) % q->max;
q->empty = 0;
q->full = q->next_insert == q->next_read;
}
pthread_cond_signal(&(q->cond));
pthread_mutex_unlock(&(q->mutex));
}