我熟悉多线程,我已经成功地用Java和Objective-C开发了许多多线程程序.但是在没有使用主线程的连接的情况下,我无法使用pthreads在C中实现以下内容:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#define NUM_OF_THREADS 2
struct thread_data {
int start;
int end;
int *arr;
};
void print(int *ints, int n);
void *processArray(void *args);
int main(int argc, const char * argv[])
{
int numOfInts = 10;
int *ints = malloc(numOfInts * sizeof(int));
for (int i = 0; i < numOfInts; i++) {
ints[i] = i;
}
print(ints, numOfInts); // prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
pthread_t threads[NUM_OF_THREADS];
struct thread_data thread_data[NUM_OF_THREADS];
// these vars are used to calculate the index ranges for each thread
int remainingWork = numOfInts, amountOfWork;
int startRange, endRange = -1;
for (int i = 0; i < NUM_OF_THREADS; i++) {
amountOfWork = remainingWork / (NUM_OF_THREADS - i);
startRange = endRange + 1;
endRange = startRange + amountOfWork - 1;
thread_data[i].arr = ints;
thread_data[i].start = startRange;
thread_data[i].end = endRange;
pthread_create(&threads[i], NULL, processArray, (void *)&thread_data[i]);
remainingWork -= amountOfWork;
}
// 1. Signal to the threads to start working
// 2. Wait for them to finish
print(ints, numOfInts); // should print [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
free(ints);
return 0;
}
void *processArray(void *args)
{
struct thread_data *data = (struct thread_data *)args;
int *arr = data->arr;
int start = data->start;
int end = data->end;
// 1. Wait for a signal to start from the main thread
for (int i = start; i <= end; i++) {
arr[i] = arr[i] + 1;
}
// 2. Signal to the main thread that you're done
pthread_exit(NULL);
}
void print(int *ints, int n)
{
printf("[");
for (int i = 0; i < n; i++) {
printf("%d", ints[i]);
if (i+1 != n)
printf(", ");
}
printf("]\n");
}
我想在上面的代码中实现以下功能:
在main()中:
在processArray()中:
我不想在主线程中使用连接,因为在实际应用程序中,主线程会创建一次线程,然后它会向后台线程发出信号多次工作,我不能让主线程线程继续,除非所有后台线程都已完成工作.在processArray函数中,我将放置一个无限循环如下:
void *processArray(void *args)
{
struct thread_data *data = (struct thread_data *)args;
while (1)
{
// 1. Wait for a signal to start from the main thread
int *arr = data->arr;
int start = data->start;
int end = data->end;
// Process
for (int i = start; i <= end; i++) {
arr[i] = arr[i] + 1;
}
// 2. Signal to the main thread that you're done
}
pthread_exit(NULL);
}
请注意,我是C和posix API的新手,请原谅我,如果我遗漏了一些明显的东西.但我真的尝试了很多东西,从使用互斥锁和一系列信号量开始,两者兼而有之,但没有成功.我认为条件变量可能会有所帮助,但我无法理解它是如何使用的.
谢谢你的时间.
问题解决了:
非常感谢你们!我终于能够安全地工作了,不按照你的提示使用连接.虽然解决方案有点难看,但它可以完成工作并且性能提升是值得的(正如您将在下面看到的那样).对于任何感兴趣的人来说,这是我正在研究的实际应用程序的模拟,其中主线程不断地为后台线程提供工作:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#define NUM_OF_THREADS 5
struct thread_data {
int id;
int start;
int end;
int *arr;
};
pthread_mutex_t currentlyIdleMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t currentlyIdleCond = PTHREAD_COND_INITIALIZER;
int currentlyIdle;
pthread_mutex_t workReadyMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t workReadyCond = PTHREAD_COND_INITIALIZER;
int workReady;
pthread_cond_t currentlyWorkingCond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t currentlyWorkingMutex= PTHREAD_MUTEX_INITIALIZER;
int currentlyWorking;
pthread_mutex_t canFinishMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t canFinishCond = PTHREAD_COND_INITIALIZER;
int canFinish;
void print(int *ints, int n);
void *processArray(void *args);
int validateResult(int *ints, int num, int start);
int main(int argc, const char * argv[])
{
int numOfInts = 10;
int *ints = malloc(numOfInts * sizeof(int));
for (int i = 0; i < numOfInts; i++) {
ints[i] = i;
}
// print(ints, numOfInts);
pthread_t threads[NUM_OF_THREADS];
struct thread_data thread_data[NUM_OF_THREADS];
workReady = 0;
canFinish = 0;
currentlyIdle = 0;
currentlyWorking = 0;
// these vars are used to calculate the index ranges for each thread
int remainingWork = numOfInts, amountOfWork;
int startRange, endRange = -1;
// Create the threads and give each one its data struct.
for (int i = 0; i < NUM_OF_THREADS; i++) {
amountOfWork = remainingWork / (NUM_OF_THREADS - i);
startRange = endRange + 1;
endRange = startRange + amountOfWork - 1;
thread_data[i].id = i;
thread_data[i].arr = ints;
thread_data[i].start = startRange;
thread_data[i].end = endRange;
pthread_create(&threads[i], NULL, processArray, (void *)&thread_data[i]);
remainingWork -= amountOfWork;
}
int loops = 1111111;
int expectedStartingValue = ints[0] + loops; // used to validate the results
// The elements in ints[] should be incremented by 1 in each loop
while (loops-- != 0) {
// Make sure all of them are ready
pthread_mutex_lock(¤tlyIdleMutex);
while (currentlyIdle != NUM_OF_THREADS) {
pthread_cond_wait(¤tlyIdleCond, ¤tlyIdleMutex);
}
pthread_mutex_unlock(¤tlyIdleMutex);
// All threads are now blocked; it's safe to not lock the mutex.
// Prevent them from finishing before authorized.
canFinish = 0;
// Reset the number of currentlyWorking threads
currentlyWorking = NUM_OF_THREADS;
// Signal to the threads to start
pthread_mutex_lock(&workReadyMutex);
workReady = 1;
pthread_cond_broadcast(&workReadyCond );
pthread_mutex_unlock(&workReadyMutex);
// Wait for them to finish
pthread_mutex_lock(¤tlyWorkingMutex);
while (currentlyWorking != 0) {
pthread_cond_wait(¤tlyWorkingCond, ¤tlyWorkingMutex);
}
pthread_mutex_unlock(¤tlyWorkingMutex);
// The threads are now waiting for permission to finish
// Prevent them from starting again
workReady = 0;
currentlyIdle = 0;
// Allow them to finish
pthread_mutex_lock(&canFinishMutex);
canFinish = 1;
pthread_cond_broadcast(&canFinishCond);
pthread_mutex_unlock(&canFinishMutex);
}
// print(ints, numOfInts);
if (validateResult(ints, numOfInts, expectedStartingValue)) {
printf("Result correct.\n");
}
else {
printf("Result invalid.\n");
}
// clean up
for (int i = 0; i < NUM_OF_THREADS; i++) {
pthread_cancel(threads[i]);
}
free(ints);
return 0;
}
void *processArray(void *args)
{
struct thread_data *data = (struct thread_data *)args;
int *arr = data->arr;
int start = data->start;
int end = data->end;
while (1) {
// Set yourself as idle and signal to the main thread, when all threads are idle main will start
pthread_mutex_lock(¤tlyIdleMutex);
currentlyIdle++;
pthread_cond_signal(¤tlyIdleCond);
pthread_mutex_unlock(¤tlyIdleMutex);
// wait for work from main
pthread_mutex_lock(&workReadyMutex);
while (!workReady) {
pthread_cond_wait(&workReadyCond , &workReadyMutex);
}
pthread_mutex_unlock(&workReadyMutex);
// Do the work
for (int i = start; i <= end; i++) {
arr[i] = arr[i] + 1;
}
// mark yourself as finished and signal to main
pthread_mutex_lock(¤tlyWorkingMutex);
currentlyWorking--;
pthread_cond_signal(¤tlyWorkingCond);
pthread_mutex_unlock(¤tlyWorkingMutex);
// Wait for permission to finish
pthread_mutex_lock(&canFinishMutex);
while (!canFinish) {
pthread_cond_wait(&canFinishCond , &canFinishMutex);
}
pthread_mutex_unlock(&canFinishMutex);
}
pthread_exit(NULL);
}
int validateResult(int *ints, int n, int start)
{
int tmp = start;
for (int i = 0; i < n; i++, tmp++) {
if (ints[i] != tmp) {
return 0;
}
}
return 1;
}
void print(int *ints, int n)
{
printf("[");
for (int i = 0; i < n; i++) {
printf("%d", ints[i]);
if (i+1 != n)
printf(", ");
}
printf("]\n");
}
我不确定是否pthread_cancel足够清理!至于障碍,如果它不仅限于@Jeremy提到的某些操作系统,它将会有很大的帮助.
基准:
我想确保这些条件实际上并没有减慢算法速度,所以我设置了这个基准来比较两个解决方案:
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <unistd.h>
#include <sys/time.h>
#include <sys/resource.h>
#define NUM_OF_THREADS 5
struct thread_data {
int start;
int end;
int *arr;
};
pthread_mutex_t currentlyIdleMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t currentlyIdleCond = PTHREAD_COND_INITIALIZER;
int currentlyIdle;
pthread_mutex_t workReadyMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t workReadyCond = PTHREAD_COND_INITIALIZER;
int workReady;
pthread_cond_t currentlyWorkingCond = PTHREAD_COND_INITIALIZER;
pthread_mutex_t currentlyWorkingMutex= PTHREAD_MUTEX_INITIALIZER;
int currentlyWorking;
pthread_mutex_t canFinishMutex = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t canFinishCond = PTHREAD_COND_INITIALIZER;
int canFinish;
void *processArrayMutex(void *args);
void *processArrayJoin(void *args);
double doItWithMutex(pthread_t *threads, struct thread_data *data, int loops);
double doItWithJoin(pthread_t *threads, struct thread_data *data, int loops);
int main(int argc, const char * argv[])
{
int numOfInts = 10;
int *join_ints = malloc(numOfInts * sizeof(int));
int *mutex_ints = malloc(numOfInts * sizeof(int));
for (int i = 0; i < numOfInts; i++) {
join_ints[i] = i;
mutex_ints[i] = i;
}
pthread_t join_threads[NUM_OF_THREADS];
pthread_t mutex_threads[NUM_OF_THREADS];
struct thread_data join_thread_data[NUM_OF_THREADS];
struct thread_data mutex_thread_data[NUM_OF_THREADS];
workReady = 0;
canFinish = 0;
currentlyIdle = 0;
currentlyWorking = 0;
int remainingWork = numOfInts, amountOfWork;
int startRange, endRange = -1;
for (int i = 0; i < NUM_OF_THREADS; i++) {
amountOfWork = remainingWork / (NUM_OF_THREADS - i);
startRange = endRange + 1;
endRange = startRange + amountOfWork - 1;
join_thread_data[i].arr = join_ints;
join_thread_data[i].start = startRange;
join_thread_data[i].end = endRange;
mutex_thread_data[i].arr = mutex_ints;
mutex_thread_data[i].start = startRange;
mutex_thread_data[i].end = endRange;
pthread_create(&mutex_threads[i], NULL, processArrayMutex, (void *)&mutex_thread_data[i]);
remainingWork -= amountOfWork;
}
int numOfBenchmarkTests = 100;
int numberOfLoopsPerTest= 1000;
double join_sum = 0.0, mutex_sum = 0.0;
for (int i = 0; i < numOfBenchmarkTests; i++)
{
double joinTime = doItWithJoin(join_threads, join_thread_data, numberOfLoopsPerTest);
double mutexTime= doItWithMutex(mutex_threads, mutex_thread_data, numberOfLoopsPerTest);
join_sum += joinTime;
mutex_sum+= mutexTime;
}
double join_avg = join_sum / numOfBenchmarkTests;
double mutex_avg= mutex_sum / numOfBenchmarkTests;
printf("Join average : %f\n", join_avg);
printf("Mutex average: %f\n", mutex_avg);
double diff = join_avg - mutex_avg;
if (diff > 0.0)
printf("Mutex is %.0f%% faster.\n", 100 * diff / join_avg);
else if (diff < 0.0)
printf("Join is %.0f%% faster.\n", 100 * diff / mutex_avg);
else
printf("Both have the same performance.");
free(join_ints);
free(mutex_ints);
return 0;
}
// From https://stackoverflow.com/a/2349941/408286
double get_time()
{
struct timeval t;
struct timezone tzp;
gettimeofday(&t, &tzp);
return t.tv_sec + t.tv_usec*1e-6;
}
double doItWithMutex(pthread_t *threads, struct thread_data *data, int num_loops)
{
double start = get_time();
int loops = num_loops;
while (loops-- != 0) {
// Make sure all of them are ready
pthread_mutex_lock(¤tlyIdleMutex);
while (currentlyIdle != NUM_OF_THREADS) {
pthread_cond_wait(¤tlyIdleCond, ¤tlyIdleMutex);
}
pthread_mutex_unlock(¤tlyIdleMutex);
// All threads are now blocked; it's safe to not lock the mutex.
// Prevent them from finishing before authorized.
canFinish = 0;
// Reset the number of currentlyWorking threads
currentlyWorking = NUM_OF_THREADS;
// Signal to the threads to start
pthread_mutex_lock(&workReadyMutex);
workReady = 1;
pthread_cond_broadcast(&workReadyCond );
pthread_mutex_unlock(&workReadyMutex);
// Wait for them to finish
pthread_mutex_lock(¤tlyWorkingMutex);
while (currentlyWorking != 0) {
pthread_cond_wait(¤tlyWorkingCond, ¤tlyWorkingMutex);
}
pthread_mutex_unlock(¤tlyWorkingMutex);
// The threads are now waiting for permission to finish
// Prevent them from starting again
workReady = 0;
currentlyIdle = 0;
// Allow them to finish
pthread_mutex_lock(&canFinishMutex);
canFinish = 1;
pthread_cond_broadcast(&canFinishCond);
pthread_mutex_unlock(&canFinishMutex);
}
return get_time() - start;
}
double doItWithJoin(pthread_t *threads, struct thread_data *data, int num_loops)
{
double start = get_time();
int loops = num_loops;
while (loops-- != 0) {
// create them
for (int i = 0; i < NUM_OF_THREADS; i++) {
pthread_create(&threads[i], NULL, processArrayJoin, (void *)&data[i]);
}
// wait
for (int i = 0; i < NUM_OF_THREADS; i++) {
pthread_join(threads[i], NULL);
}
}
return get_time() - start;
}
void *processArrayMutex(void *args)
{
struct thread_data *data = (struct thread_data *)args;
int *arr = data->arr;
int start = data->start;
int end = data->end;
while (1) {
// Set yourself as idle and signal to the main thread, when all threads are idle main will start
pthread_mutex_lock(¤tlyIdleMutex);
currentlyIdle++;
pthread_cond_signal(¤tlyIdleCond);
pthread_mutex_unlock(¤tlyIdleMutex);
// wait for work from main
pthread_mutex_lock(&workReadyMutex);
while (!workReady) {
pthread_cond_wait(&workReadyCond , &workReadyMutex);
}
pthread_mutex_unlock(&workReadyMutex);
// Do the work
for (int i = start; i <= end; i++) {
arr[i] = arr[i] + 1;
}
// mark yourself as finished and signal to main
pthread_mutex_lock(¤tlyWorkingMutex);
currentlyWorking--;
pthread_cond_signal(¤tlyWorkingCond);
pthread_mutex_unlock(¤tlyWorkingMutex);
// Wait for permission to finish
pthread_mutex_lock(&canFinishMutex);
while (!canFinish) {
pthread_cond_wait(&canFinishCond , &canFinishMutex);
}
pthread_mutex_unlock(&canFinishMutex);
}
pthread_exit(NULL);
}
void *processArrayJoin(void *args)
{
struct thread_data *data = (struct thread_data *)args;
int *arr = data->arr;
int start = data->start;
int end = data->end;
// Do the work
for (int i = start; i <= end; i++) {
arr[i] = arr[i] + 1;
}
pthread_exit(NULL);
}
输出是:
Join average : 0.153074
Mutex average: 0.071588
Mutex is 53% faster.
再次感谢你.我真的很感谢你的帮助!
join很明显，您需要使用与 不同的同步技术。
不幸的是你有很多选择。一个是“同步屏障”，基本上是每个到达它的线程都会阻塞，直到它们全部到达它（您提前指定线程数）。看着pthread_barrier。
另一种方法是使用条件变量/互斥体对 ( pthread_cond_*)。当每个线程完成时，它会获取互斥锁，增加计数，并向 condvar 发出信号。主线程等待 condvar，直到计数达到它期望的值。代码如下所示：
// thread has finished
mutex_lock
++global_count
// optional optimization: only execute the next line when global_count >= N
cond_signal
mutex_unlock
// main is waiting for N threads to finish
mutex_lock
while (global_count < N) {
cond_wait
}
mutex_unlock
另一种方法是每个线程使用一个信号量——当线程完成时，它会发布自己的信号量，并且主线程依次等待每个信号量，而不是依次加入每个线程。
您还需要同步来重新启动下一个作业的线程——这可能是与第一个同步对象类型相同的第二个同步对象，详细信息已更改，因为您有 1 个海报和 N 个等待者，而不是其他方式大约。或者您可以（小心地）将同一个对象重复用于这两个目的。
如果您已经尝试过这些操作，但您的代码不起作用，也许可以针对您尝试的代码提出一个新的具体问题。所有这些都足以胜任这项任务。