无法重新获取互斥量并在线程之间正确传递值
unable to reacquire mutex and pass values correctly between threads
我正在尝试实现一个代码来练习同步，因此可能不是最佳设计或方法，但目标如下
主线程
创建 100 个整数的负载并等待任何线程可用
当它从一个可用的线程获得信号时 - 它会解锁用于复制的有效载荷并继续创建另一个有效载荷
工作线程
在创建它时使其自身可用于数据处理并发送其可用的信号
尝试从主线程锁定数据负载并将其复制到本地数组
（在这里观察错误 - 无法正确访问数据）
关闭可用标志
（无法将可用状态关闭）
通过本地副本继续处理数据
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <stdbool.h>
#define WORKERS 2
#define ARRAY_ELEMENTS 100
#define MAX 1000
pthread_mutex_t mutex_bucket1 = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t mutex_signal = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_go = PTHREAD_COND_INITIALIZER;
pthread_cond_t cond_busy = PTHREAD_COND_INITIALIZER;
static int value = 0;
bool available = false;
void *worker_thread(void *pbucket)
{
sleep(5);
while(1)
{
unsigned int count = 0;
int local_array[ARRAY_ELEMENTS];
int *ptbucket = (int*)pbucket;
setbuf(stdout, NULL);
pthread_mutex_lock(&mutex_signal);
printf(" -------------- \n chainging state to available \n --------- ");
available = true;
printf(" -------------- \n from thread sending go signal \n --------- ");
pthread_cond_signal(&cond_go);
pthread_mutex_unlock(&mutex_signal);
pthread_mutex_lock(&mutex_bucket1);
printf(" -------------- \n data part locked in thread for copying \n --------- ");
while(count < ARRAY_ELEMENTS)
{
printf(" %d - \n", ptbucket[count]); /***incorrect values***/
local_array[count] = ptbucket[count];
count++;
}
pthread_mutex_unlock(&mutex_bucket1);
/*Never able to acquire mutex_signal and change state to not available*/ **BUG**
pthread_mutex_lock(&mutex_signal);
printf(" -------------- \n chainging state to not available \n --------- ");
available = false;
pthread_mutex_unlock(&mutex_signal);
count = 0;
while(count < ARRAY_ELEMENTS)
{
printf(" %d - \n", local_array[count]);
count++;
}
printf(" -------------- \n about to sleep for 5secs \n --------- ");
sleep(5);
}
}
int main(void)
{
pthread_t thread_id[WORKERS];
unsigned int* pbucket1 = (int*) malloc(sizeof(int) * ARRAY_ELEMENTS);
unsigned int* pbucket;
for(int i = 0; i < WORKERS - 1; i++)
{
pthread_create(&thread_id[i], NULL, worker_thread, (void *) pbucket);
}
for(int i = 0; i < MAX; i++)
{
unsigned int count = 0;
pbucket = pbucket1;
// Make the payload ready
pthread_mutex_lock(&mutex_bucket1);
printf(" -------------- creating data payload --------- \n");
while(count < ARRAY_ELEMENTS)
{
pbucket1[count] = i;
i++;
count++;
}
printf(" -------------- \n waiting for go signal \n --------- ");
while(!available)
{
pthread_cond_wait(&cond_go, &mutex_signal);
}
pthread_mutex_unlock(&mutex_bucket1);
/*I believe after we unlock variable "available" can be mutexed
again by other thread but seems thinking is flawed */
printf(" -------------- \n Main thread sleep for 3 seconds \n --------- ");
sleep(3);
}
for(int i = 0; i < WORKERS; i++)
{
pthread_join(thread_id[i], NULL);
}
return 0;
}
我认为你的某些想法是倒退的；等待的不应该是主上下文，应该是等待数据的工作线程...
主线程的工作应该是不断填充负载并一次唤醒一个线程来处理它。
所以这里有一些潦草的代码，我认为更明智一些：
/**
file: answer.c
compile: gcc -o answer answer.c -pthread
usage: answer [numThreads] [numElements]
**/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#define STATE_WAIT 1
#define STATE_READY 2
void *routine(void*);
typedef struct _shared_t {
pthread_mutex_t m;
pthread_cond_t c;
unsigned char state;
int *payload;
size_t numElements;
pthread_t *threads;
size_t numThreads;
} shared_t;
static inline void shared_init(shared_t *shared, size_t numThreads, size_t numElements) {
memset(shared, 0, sizeof(shared_t));
pthread_mutex_init(&shared->m, NULL);
pthread_cond_init(&shared->c, NULL);
shared->state = STATE_WAIT;
shared->numThreads = numThreads;
shared->numElements = numElements;
{
int it = 0;
shared->threads = (pthread_t*) calloc(shared->numThreads, sizeof(pthread_t));
while (it < shared->numThreads) {
if (pthread_create(&shared->threads[it], NULL, routine, shared) != 0) {
break;
}
it++;
}
}
}
static inline void shared_populate(shared_t *shared) {
if (pthread_mutex_lock(&shared->m) != 0) {
return;
}
shared->payload = (int*) calloc(shared->numElements, sizeof(int));
{
int it = 0,
end = shared->numElements;
while (it < end) {
shared->payload[it] = rand();
it++;
}
}
shared->state = STATE_READY;
pthread_cond_signal(&shared->c);
pthread_mutex_unlock(&shared->m);
}
static inline void shared_cleanup(shared_t *shared) {
int it = 0,
end = shared->numThreads;
while (it < end) {
pthread_join(shared->threads[it], NULL);
}
pthread_mutex_destroy(&shared->m);
pthread_cond_destroy(&shared->c);
free(shared->threads);
}
void* routine(void *arg) {
shared_t *shared = (shared_t*) arg;
int *payload;
do {
if (pthread_mutex_lock(&shared->m) != 0) {
break;
}
while (shared->state == STATE_WAIT) {
pthread_cond_wait(&shared->c, &shared->m);
}
payload = shared->payload;
shared->state = STATE_WAIT;
pthread_mutex_unlock(&shared->m);
if (payload) {
int it = 0,
end = shared->numElements;
while (it < end) {
printf("Thread #%ld got payload %p(%d)=%d\n",
pthread_self(), payload, it, payload[it]);
it++;
}
free(payload);
}
} while(1);
pthread_exit(NULL);
}
int main(int argc, char *argv[]) {
shared_t shared;
int numThreads = argc > 1 ? atoi(argv[1]) : 1;
int numElements = argc > 2 ? atoi(argv[2]) : 100;
shared_init(&shared, numThreads, numElements);
do {
shared_populate(&shared);
} while (1);
shared_cleanup(&shared);
return 0;
}
显然，上面的代码对错误的容忍度不高，并且不容易干净地关闭...仅供说明。
我们先来看一下main，这样我们就知道主程序的流程是怎样的：
int main(int argc, char *argv[]) {
shared_t shared;
int numThreads = argc > 1 ? atoi(argv[1]) : 1;
int numElements = argc > 2 ? atoi(argv[2]) : 100;
shared_init(&shared, numThreads, numElements);
do {
shared_populate(&shared);
} while (1);
shared_cleanup(&shared);
return 0;
}
它在堆栈上保留 shared_t：
typedef struct _shared_t {
pthread_mutex_t m;
pthread_cond_t c;
unsigned char state;
int *payload;
size_t numElements;
pthread_t *threads;
size_t numThreads;
} shared_t;
同步需要大部分自我解释、互斥、条件和状态。
首先 shared_t 必须使用提供的选项用互斥体、条件、状态和线程进行初始化：
static inline void shared_init(shared_t *shared, size_t numThreads, size_t numElements) {
memset(shared, 0, sizeof(shared_t));
pthread_mutex_init(&shared->m, NULL);
pthread_cond_init(&shared->c, NULL);
shared->state = STATE_WAIT;
shared->numThreads = numThreads;
shared->numElements = numElements;
{
int it = 0;
shared->threads = (pthread_t*) calloc(shared->numThreads, sizeof(pthread_t));
while (it < shared->numThreads) {
if (pthread_create(&shared->threads[it], NULL, routine, shared) != 0) {
break;
}
it++;
}
}
}
当工作线程被这个例程创建时，它们被强制进入等待状态。
循环中对shared_populate的第一次调用在将负载设置为一些随机数后唤醒第一个线程：
static inline void shared_populate(shared_t *shared) {
if (pthread_mutex_lock(&shared->m) != 0) {
return;
}
shared->payload = (int*) calloc(shared->numElements, sizeof(int));
{
int it = 0,
end = shared->numElements;
while (it < end) {
shared->payload[it] = rand();
it++;
}
}
shared->state = STATE_READY;
pthread_cond_signal(&shared->c);
pthread_mutex_unlock(&shared->m);
}
注意 pthread_cond_signal 而不是 pthread_cond_broadcast，因为我们只想唤醒第一个线程。
void* routine(void *arg) {
shared_t *shared = (shared_t*) arg;
int *payload;
do {
if (pthread_mutex_lock(&shared->m) != 0) {
break;
}
while (shared->state == STATE_WAIT) {
pthread_cond_wait(&shared->c, &shared->m);
}
payload = shared->payload;
shared->state = STATE_WAIT;
pthread_mutex_unlock(&shared->m);
if (payload) {
int it = 0,
end = shared->numElements;
while (it < end) {
printf("Thread #%ld got payload %p(%d)=%d\n",
pthread_self(), payload, it, payload[it]);
it++;
}
free(payload);
}
} while(1);
pthread_exit(NULL);
}
所以我们在routine调用pthread_cond_wait时醒来，状态已经改变，所以我们跳出循环，我们保存指向payload的指针，将状态重置为等待，然后释放互斥量。
此时main可以重新填充payload并唤醒下一个线程，同时当前工作线程可以处理，然后释放payload。
一些建议：
- 始终使用尽可能少的互斥锁和条件变量 (KISS)
- 研究条件变量的原子性
- 始终遵循有关获取和释放互斥量以及条件变量信号的基本规则：
- 如果您锁定了它，请将其解锁。
- 永远只等待 某些东西：绝对需要预测等待循环，一直都是。
如果你不能重现我所做的，那就拿出代码并尝试对其进行扩展；您需要做的第一件事是能够正常关闭进程（输入 shared_cleanup），也许您需要可变大小的有效负载，或者原始问题中未提及的其他一些要求。
关于 printf 的注意事项 ... 不能保证附加到流中是原子的，碰巧大多数时候在 *nix 上都是 ... 因为我们是只是做展示和讲述，我们不需要关心这个......通常，不要依赖任何流操作的原子性......
我正在尝试实现一个代码来练习同步，因此可能不是最佳设计或方法，但目标如下
主线程
创建 100 个整数的负载并等待任何线程可用
当它从一个可用的线程获得信号时 - 它会解锁用于复制的有效载荷并继续创建另一个有效载荷
工作线程
在创建它时使其自身可用于数据处理并发送其可用的信号
尝试从主线程锁定数据负载并将其复制到本地数组 （在这里观察错误 - 无法正确访问数据）
关闭可用标志 （无法将可用状态关闭）
通过本地副本继续处理数据
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <stdbool.h>
#define WORKERS 2
#define ARRAY_ELEMENTS 100
#define MAX 1000
pthread_mutex_t mutex_bucket1 = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t mutex_signal = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_go = PTHREAD_COND_INITIALIZER;
pthread_cond_t cond_busy = PTHREAD_COND_INITIALIZER;
static int value = 0;
bool available = false;
void *worker_thread(void *pbucket)
{
sleep(5);
while(1)
{
unsigned int count = 0;
int local_array[ARRAY_ELEMENTS];
int *ptbucket = (int*)pbucket;
setbuf(stdout, NULL);
pthread_mutex_lock(&mutex_signal);
printf(" -------------- \n chainging state to available \n --------- ");
available = true;
printf(" -------------- \n from thread sending go signal \n --------- ");
pthread_cond_signal(&cond_go);
pthread_mutex_unlock(&mutex_signal);
pthread_mutex_lock(&mutex_bucket1);
printf(" -------------- \n data part locked in thread for copying \n --------- ");
while(count < ARRAY_ELEMENTS)
{
printf(" %d - \n", ptbucket[count]); /***incorrect values***/
local_array[count] = ptbucket[count];
count++;
}
pthread_mutex_unlock(&mutex_bucket1);
/*Never able to acquire mutex_signal and change state to not available*/ **BUG**
pthread_mutex_lock(&mutex_signal);
printf(" -------------- \n chainging state to not available \n --------- ");
available = false;
pthread_mutex_unlock(&mutex_signal);
count = 0;
while(count < ARRAY_ELEMENTS)
{
printf(" %d - \n", local_array[count]);
count++;
}
printf(" -------------- \n about to sleep for 5secs \n --------- ");
sleep(5);
}
}
int main(void)
{
pthread_t thread_id[WORKERS];
unsigned int* pbucket1 = (int*) malloc(sizeof(int) * ARRAY_ELEMENTS);
unsigned int* pbucket;
for(int i = 0; i < WORKERS - 1; i++)
{
pthread_create(&thread_id[i], NULL, worker_thread, (void *) pbucket);
}
for(int i = 0; i < MAX; i++)
{
unsigned int count = 0;
pbucket = pbucket1;
// Make the payload ready
pthread_mutex_lock(&mutex_bucket1);
printf(" -------------- creating data payload --------- \n");
while(count < ARRAY_ELEMENTS)
{
pbucket1[count] = i;
i++;
count++;
}
printf(" -------------- \n waiting for go signal \n --------- ");
while(!available)
{
pthread_cond_wait(&cond_go, &mutex_signal);
}
pthread_mutex_unlock(&mutex_bucket1);
/*I believe after we unlock variable "available" can be mutexed
again by other thread but seems thinking is flawed */
printf(" -------------- \n Main thread sleep for 3 seconds \n --------- ");
sleep(3);
}
for(int i = 0; i < WORKERS; i++)
{
pthread_join(thread_id[i], NULL);
}
return 0;
}
我认为你的某些想法是倒退的；等待的不应该是主上下文，应该是等待数据的工作线程...
主线程的工作应该是不断填充负载并一次唤醒一个线程来处理它。
所以这里有一些潦草的代码，我认为更明智一些：
/**
file: answer.c
compile: gcc -o answer answer.c -pthread
usage: answer [numThreads] [numElements]
**/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <pthread.h>
#define STATE_WAIT 1
#define STATE_READY 2
void *routine(void*);
typedef struct _shared_t {
pthread_mutex_t m;
pthread_cond_t c;
unsigned char state;
int *payload;
size_t numElements;
pthread_t *threads;
size_t numThreads;
} shared_t;
static inline void shared_init(shared_t *shared, size_t numThreads, size_t numElements) {
memset(shared, 0, sizeof(shared_t));
pthread_mutex_init(&shared->m, NULL);
pthread_cond_init(&shared->c, NULL);
shared->state = STATE_WAIT;
shared->numThreads = numThreads;
shared->numElements = numElements;
{
int it = 0;
shared->threads = (pthread_t*) calloc(shared->numThreads, sizeof(pthread_t));
while (it < shared->numThreads) {
if (pthread_create(&shared->threads[it], NULL, routine, shared) != 0) {
break;
}
it++;
}
}
}
static inline void shared_populate(shared_t *shared) {
if (pthread_mutex_lock(&shared->m) != 0) {
return;
}
shared->payload = (int*) calloc(shared->numElements, sizeof(int));
{
int it = 0,
end = shared->numElements;
while (it < end) {
shared->payload[it] = rand();
it++;
}
}
shared->state = STATE_READY;
pthread_cond_signal(&shared->c);
pthread_mutex_unlock(&shared->m);
}
static inline void shared_cleanup(shared_t *shared) {
int it = 0,
end = shared->numThreads;
while (it < end) {
pthread_join(shared->threads[it], NULL);
}
pthread_mutex_destroy(&shared->m);
pthread_cond_destroy(&shared->c);
free(shared->threads);
}
void* routine(void *arg) {
shared_t *shared = (shared_t*) arg;
int *payload;
do {
if (pthread_mutex_lock(&shared->m) != 0) {
break;
}
while (shared->state == STATE_WAIT) {
pthread_cond_wait(&shared->c, &shared->m);
}
payload = shared->payload;
shared->state = STATE_WAIT;
pthread_mutex_unlock(&shared->m);
if (payload) {
int it = 0,
end = shared->numElements;
while (it < end) {
printf("Thread #%ld got payload %p(%d)=%d\n",
pthread_self(), payload, it, payload[it]);
it++;
}
free(payload);
}
} while(1);
pthread_exit(NULL);
}
int main(int argc, char *argv[]) {
shared_t shared;
int numThreads = argc > 1 ? atoi(argv[1]) : 1;
int numElements = argc > 2 ? atoi(argv[2]) : 100;
shared_init(&shared, numThreads, numElements);
do {
shared_populate(&shared);
} while (1);
shared_cleanup(&shared);
return 0;
}
显然，上面的代码对错误的容忍度不高，并且不容易干净地关闭...仅供说明。
我们先来看一下main，这样我们就知道主程序的流程是怎样的：
int main(int argc, char *argv[]) {
shared_t shared;
int numThreads = argc > 1 ? atoi(argv[1]) : 1;
int numElements = argc > 2 ? atoi(argv[2]) : 100;
shared_init(&shared, numThreads, numElements);
do {
shared_populate(&shared);
} while (1);
shared_cleanup(&shared);
return 0;
}
它在堆栈上保留 shared_t：
typedef struct _shared_t {
pthread_mutex_t m;
pthread_cond_t c;
unsigned char state;
int *payload;
size_t numElements;
pthread_t *threads;
size_t numThreads;
} shared_t;
同步需要大部分自我解释、互斥、条件和状态。
首先 shared_t 必须使用提供的选项用互斥体、条件、状态和线程进行初始化：
static inline void shared_init(shared_t *shared, size_t numThreads, size_t numElements) {
memset(shared, 0, sizeof(shared_t));
pthread_mutex_init(&shared->m, NULL);
pthread_cond_init(&shared->c, NULL);
shared->state = STATE_WAIT;
shared->numThreads = numThreads;
shared->numElements = numElements;
{
int it = 0;
shared->threads = (pthread_t*) calloc(shared->numThreads, sizeof(pthread_t));
while (it < shared->numThreads) {
if (pthread_create(&shared->threads[it], NULL, routine, shared) != 0) {
break;
}
it++;
}
}
}
当工作线程被这个例程创建时，它们被强制进入等待状态。
循环中对shared_populate的第一次调用在将负载设置为一些随机数后唤醒第一个线程：
static inline void shared_populate(shared_t *shared) {
if (pthread_mutex_lock(&shared->m) != 0) {
return;
}
shared->payload = (int*) calloc(shared->numElements, sizeof(int));
{
int it = 0,
end = shared->numElements;
while (it < end) {
shared->payload[it] = rand();
it++;
}
}
shared->state = STATE_READY;
pthread_cond_signal(&shared->c);
pthread_mutex_unlock(&shared->m);
}
注意 pthread_cond_signal 而不是 pthread_cond_broadcast，因为我们只想唤醒第一个线程。
void* routine(void *arg) {
shared_t *shared = (shared_t*) arg;
int *payload;
do {
if (pthread_mutex_lock(&shared->m) != 0) {
break;
}
while (shared->state == STATE_WAIT) {
pthread_cond_wait(&shared->c, &shared->m);
}
payload = shared->payload;
shared->state = STATE_WAIT;
pthread_mutex_unlock(&shared->m);
if (payload) {
int it = 0,
end = shared->numElements;
while (it < end) {
printf("Thread #%ld got payload %p(%d)=%d\n",
pthread_self(), payload, it, payload[it]);
it++;
}
free(payload);
}
} while(1);
pthread_exit(NULL);
}
所以我们在routine调用pthread_cond_wait时醒来，状态已经改变，所以我们跳出循环，我们保存指向payload的指针，将状态重置为等待，然后释放互斥量。
此时main可以重新填充payload并唤醒下一个线程，同时当前工作线程可以处理，然后释放payload。
一些建议：
- 始终使用尽可能少的互斥锁和条件变量 (KISS)
- 研究条件变量的原子性
- 始终遵循有关获取和释放互斥量以及条件变量信号的基本规则：
- 如果您锁定了它，请将其解锁。
- 永远只等待 某些东西：绝对需要预测等待循环，一直都是。
如果你不能重现我所做的，那就拿出代码并尝试对其进行扩展；您需要做的第一件事是能够正常关闭进程（输入 shared_cleanup），也许您需要可变大小的有效负载，或者原始问题中未提及的其他一些要求。
关于 printf 的注意事项 ... 不能保证附加到流中是原子的，碰巧大多数时候在 *nix 上都是 ... 因为我们是只是做展示和讲述，我们不需要关心这个......通常，不要依赖任何流操作的原子性......