我是多线程的新手，我正在尝试实现一个简单的线程安全任务队列，其中每个线程都可以完成工作，直到不再有任务为止。任何线程都不会排队任务。
为了测试目的，每个任务只包含一个数字。
static pthread_mutex_t task_mutex = PTHREAD_MUTEX_INITIALIZER;
typedef struct Task{
int number;
}Task;
typedef struct Cell{
Task t;
struct Cell* next;
}Cell;
typedef struct TQueue{
struct Cell* head;
struct Cell* tail;
}TQueue;
int empty(TQueue *Queue)
return queue->head == queue->tail;
void startQueue(TQueue *queue){
queue->head = malloc(sizeof(Cell));
queue->tail = queue->head;
}
void enqueue(TQueue *queue, Task C){
queue->tail->next = malloc(sizeof(Cell));
queue->tail = queue->tail->next;
queue->tail->t = C;
queue->tail->next = NULL;
}
Task * dequeue(TQueue* queue){
pthread_mutex_lock( &task_mutex);
Task * t;
if(empty(queue)) t = NULL;
else{
struct Cell* p = queue->head;
queue->head = queue->head->next;
t = &queue->head->t;
free(p);
}
pthread_mutex_unlock( &task_mutex);
return t;
}
void * work( void* arg){
TQueue* queue = (TQueue *)arg;
Task* t = malloc(sizeof(Task));
for(t = dequeue(queue); t != NULL; t = dequeue(queue))
printf("%d ", t->number);
free(t);
pthread_exit(NULL);
return 0;
}
对于一个简单的测试，我在main上运行了这个：
int main(){
TQueue* queue = malloc(sizeof(TQueue));
startQueue(queue);
pthread_t threads[3];
Task t[3];
for(int i = 0; i < 3; i++){
t[i].number = i + 1;
enqueue(queue, t[i]);
}
for(int i = 0; i < 3; i++) pthread_create(&threads[i], NULL, work, (void*)queue);
for(int i = 0; i < 3; i++) pthread_join(threads[i], NULL);
return 0;
}
任何顺序的预期输出都是1 2 3，但有时会打印出一个奇怪数字的序列，如1823219 2 3。我无法发现任何竞争条件或相关问题，所以我感谢任何帮助。
答案 0 :(得分：0)
我发现了一些错误。
我已经注释了您的代码。我从你的第一个帖子和你的第二个帖子中取了一点。我修改了代码，显示前后[请原谅无偿风格的清理]：
#include <stdio.h>
#include <pthread.h>
#include <malloc.h>
static pthread_mutex_t task_mutex = PTHREAD_MUTEX_INITIALIZER;
typedef struct Task {
int number;
} Task;
typedef struct Cell {
// NOTE/BUG: this should be a pointer to the task. otherwise, dequeue gets
// messy
#if 0
Task t;
#else
Task *t;
#endif
struct Cell *next;
} Cell;
typedef struct TQueue {
struct Cell *head;
struct Cell *tail;
} TQueue;
void
startQueue(TQueue *queue)
{
#if 0
queue->head = malloc(sizeof(Cell));
#else
queue->head = NULL;
#endif
queue->tail = NULL;
}
int
empty(TQueue *queue)
{
// NOTE/BUG: dequeue never touches tail, so this test is incorrect
#if 0
return (queue->head == queue->tail);
#else
return (queue->head == NULL);
#endif
}
void
enqueue(TQueue *queue, Task *t)
{
Cell *p;
pthread_mutex_lock(&task_mutex);
p = malloc(sizeof(Cell));
p->next = NULL;
p->t = t;
if (queue->tail == NULL) {
queue->tail = p;
queue->head = p;
}
else {
queue->tail->next = p;
queue->tail = p;
}
pthread_mutex_unlock(&task_mutex);
}
Task *
dequeue(TQueue *queue)
{
Task *t;
pthread_mutex_lock(&task_mutex);
if (empty(queue))
t = NULL;
else {
Cell *p = queue->head;
if (p == queue->tail)
queue->tail = NULL;
queue->head = p->next;
// NOTE/BUG: this is setting t to the second element in the list,
// not the first
// NOTE/BUG: this is also undefined behavior, in original code (with
// original struct definition), because what t points to _does_ get
// freed before return
#if 0
t = &queue->head->t;
#else
t = p->t;
#endif
free(p);
}
pthread_mutex_unlock(&task_mutex);
return t;
}
void *
work(void *arg)
{
TQueue *queue = (TQueue *) arg;
// NOTE/BUG: this gets orphaned on the first call to dequeue
#if 0
Task *t = malloc(sizeof(Task));
#else
Task *t;
#endif
for (t = dequeue(queue); t != NULL; t = dequeue(queue))
printf("%d ", t->number);
// NOTE/BUG: this frees some cell allocated in main -- not what we want
#if 0
free(t);
#endif
pthread_exit(NULL);
return 0;
}
// For a simple test i runned this on main:
int
main()
{
TQueue *queue = malloc(sizeof(TQueue));
startQueue(queue);
pthread_t threads[3];
Task t[3];
for (int i = 0; i < 3; i++) {
t[i].number = i + 1;
#if 0
enqueue(queue, t);
#else
enqueue(queue, &t[i]);
#endif
}
for (int i = 0; i < 3; i++)
pthread_create(&threads[i], NULL, work, (void *) queue);
for (int i = 0; i < 3; i++)
pthread_join(threads[i], NULL);
return 0;
}
<强>更新
线程是否同时执行任务？我已经用htop测试了cpu的使用情况，我只能最多使用四个核心的单个核心。
要记住一些事项。 htop可能不会对运行时间短的程序表现出色。即使有10,000个队列条目，该程序也会在20ms内执行。
让程序本身打印信息更好[见下文]。请注意，printf会对stdin进行线程锁定，因此它可能会对＆＃34; serial＆＃34;该计划的性质。它还为程序的执行时间贡献重要量（即printf比dequeue慢得多
此外，一个线程（即第一个线程）可以独占队列并在其他人有机会运行之前排除所有条目。
操作系统可以[自由地]安排单个核心上的所有线程。然后它可以＆＃34;迁移＆＃34;他们以后（例如在一秒钟左右）。
我已经增强了该程序，在输出打印中包含一些时间信息，可能有助于显示您希望看到的更多信息。此外，我添加了命令行选项来控制线程数和排队的项目数。这与我自己的一些程序类似。将程序输出转移到日志文件并检查它。在多次运行中使用选项
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <malloc.h>
#include <time.h>
int opt_n; // suppress thread output
int opt_T; // number of threads
int opt_Q; // number of queue items
static pthread_mutex_t task_mutex = PTHREAD_MUTEX_INITIALIZER;
double tvzero;
typedef struct Task {
int number;
} Task;
typedef struct Cell {
Task *t;
struct Cell *next;
} Cell;
typedef struct TQueue {
struct Cell *head;
struct Cell *tail;
} TQueue;
typedef struct Thread {
pthread_t tid;
int xid;
TQueue *queue;
} Thread;
double
tvgetf(void)
{
struct timespec ts;
double sec;
clock_gettime(CLOCK_REALTIME,&ts);
sec = ts.tv_nsec;
sec /= 1e9;
sec += ts.tv_sec;
sec -= tvzero;
return sec;
}
void
startQueue(TQueue *queue)
{
queue->head = NULL;
queue->tail = NULL;
}
int
empty(TQueue *queue)
{
return (queue->head == NULL);
}
void
enqueue(TQueue *queue, Task *t)
{
Cell *p;
pthread_mutex_lock(&task_mutex);
p = malloc(sizeof(Cell));
p->next = NULL;
p->t = t;
if (queue->tail == NULL) {
queue->tail = p;
queue->head = p;
}
else {
queue->tail->next = p;
queue->tail = p;
}
pthread_mutex_unlock(&task_mutex);
}
Task *
dequeue(TQueue *queue)
{
Task *t;
pthread_mutex_lock(&task_mutex);
if (empty(queue))
t = NULL;
else {
Cell *p = queue->head;
if (p == queue->tail)
queue->tail = NULL;
queue->head = p->next;
t = p->t;
free(p);
}
pthread_mutex_unlock(&task_mutex);
return t;
}
void *
work(void *arg)
{
Thread *tskcur = arg;
TQueue *queue = tskcur->queue;
Task *t;
double tvbef;
double tvaft;
while (1) {
tvbef = tvgetf();
t = dequeue(queue);
tvaft = tvgetf();
if (t == NULL)
break;
if (! opt_n)
printf("[%.9f/%.9f %5.5d] %d\n",
tvbef,tvaft - tvbef,tskcur->xid,t->number);
}
return (void *) 0;
}
// For a simple test i runned this on main:
int
main(int argc,char **argv)
{
char *cp;
TQueue *queue;
Task *t;
Thread *tsk;
--argc;
++argv;
for (; argc > 0; --argc, ++argv) {
cp = *argv;
if (*cp != '-')
break;
switch (cp[1]) {
case 'n': // suppress thread output
opt_n = 1;
break;
case 'Q': // number of queue items
opt_Q = atoi(cp + 2);
break;
case 'T': // number of threads
opt_T = atoi(cp + 2);
break;
default:
break;
}
}
tvzero = tvgetf();
queue = malloc(sizeof(TQueue));
startQueue(queue);
if (opt_T == 0)
opt_T = 16;
Thread threads[opt_T];
if (opt_Q == 0)
opt_Q = 10000;
t = malloc(sizeof(Task) * opt_Q);
for (int i = 0; i < opt_Q; i++) {
t[i].number = i + 1;
enqueue(queue, &t[i]);
}
for (int i = 0; i < opt_T; i++) {
tsk = &threads[i];
tsk->xid = i + 1;
tsk->queue = queue;
pthread_create(&tsk->tid, NULL, work, tsk);
}
for (int i = 0; i < opt_T; i++) {
tsk = &threads[i];
pthread_join(tsk->tid, NULL);
}
printf("TOTAL: %.9f\n",tvgetf());
free(t);
return 0;
}
更新＃2：
此外，一个线程（即第一个线程）可以独占队列并在其他人有机会运行之前排空所有条目。＆＃34;在那种情况下可以做些什么？
一些事情。
pthread_create需要一些时间，允许线程1继续，而其他线程仍在创建中。改善这种情况的一种方法是创建所有线程，每个线程设置一个＆＃34;我正在运行＆＃34; flag（在其线程控制块中）。主线程等待所有线程设置此标志。然后，主线程设置一个全局volatile＆＃34; you_may_now_all_run＆＃34;标记每个线程在进入其主线程循环之前旋转。根据我的经验，它们都会在彼此之间的微秒内开始运行[或更好]。
我在下面的更新代码中没有实现这一点，因此您可以自己尝试[以及nanosleep]。
互斥体整体上相当公平[至少在linux下]，因为被阻塞的线程将排队，等待互斥锁。正如我在评论中提到的那样，也可以使用nanosleep，但这[有点]会失败，因为线程会变慢。
线索饥饿的解药是＆＃34;公平＆＃34;。正如我所提到的，有一个精细的公平算法，无需等待。这是Kogan / Petrank算法：http://www.cs.technion.ac.il/~erez/Papers/wf-methodology-ppopp12.pdf这实际上有点涉及/进阶，所以需要注意......
但是，妥协可能是门票锁定：https://en.wikipedia.org/wiki/Ticket_lock
我再次重新编写了该程序。它具有池化分配，票证与互斥锁定以及日志条目的延迟打印的选项。它还交叉检查线程之间的结果，以确保它们都没有重复的条目。
当然，所有这一切的关键是准确，高精度的记录（即如果你不能测量它，你就不能调整它）。
例如，有人会认为在free内执行dequeue比简单地将Cell释放到可恢复池（类似于slab分配器）要慢，但是，性能提升不是＆＃39 ; t和预期一样伟大。这可能是因为glibc的malloc/free速度非常快[这是他们声称的]。
这些不同版本应该为您提供有关如何构建自己的性能度量套件的一些想法。
无论如何，这里的代码是：
#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <stdatomic.h>
#include <malloc.h>
#include <errno.h>
#include <string.h>
#include <time.h>
int opt_p; // print thread output immediately
int opt_T; // number of threads
int opt_Q; // number of queue items
int opt_L; // use ticket lock
int opt_M; // use fast cell alloc/free
typedef unsigned char byte;
typedef unsigned int u32;
#define sysfault(_fmt...) \
do { \
fprintf(stderr,_fmt); \
exit(1); \
} while (0)
// lock control
typedef struct AnyLock {
pthread_mutex_t mutex; // standard mutex
volatile u32 seqreq; // ticket lock request
volatile u32 seqacq; // ticket lock grant
} AnyLock;
// work value
typedef struct Task {
union {
struct Task *next;
int number;
};
} Task;
// queue item
typedef struct Cell {
struct Cell *next;
Task *t;
} Cell;
// queue control
typedef struct TQueue {
struct Cell *head;
struct Cell *tail;
} TQueue;
// thread log entry
typedef struct Log {
double tvbef;
double tvaft;
int number;
} Log;
#define BTVOFF(_off) \
((_off) >> 3)
#define BTVMSK(_off) \
(1u << ((_off) & 0x07))
#define BTVLEN(_len) \
((_len) + 7) >> 3
// thread control
typedef struct Thread {
pthread_t tid;
int xid;
TQueue *queue;
Log *log;
byte *bitv;
} Thread;
static inline byte
btvset(byte *bitv,long off)
{
u32 msk;
byte oval;
bitv += BTVOFF(off);
msk = BTVMSK(off);
oval = *bitv & msk;
*bitv |= msk;
return oval;
}
AnyLock task_mutex;
AnyLock print_mutex;
double tvzero;
Cell *cellpool; // free pool of cells
long bitvlen;
#define BARRIER \
__asm__ __volatile__("" ::: "memory")
// virtual function pointers
Cell *(*cellnew)(void);
void (*cellfree)(Cell *);
void (*lock_acquire)(AnyLock *lock);
void (*lock_release)(AnyLock *lock);
double
tvgetf(void)
{
struct timespec ts;
double sec;
clock_gettime(CLOCK_REALTIME,&ts);
sec = ts.tv_nsec;
sec /= 1e9;
sec += ts.tv_sec;
sec -= tvzero;
return sec;
}
void *
xalloc(size_t cnt,size_t siz)
{
void *ptr;
ptr = calloc(cnt,siz);
if (ptr == NULL)
sysfault("xalloc: calloc failure -- %s\n",strerror(errno));
return ptr;
}
void
lock_wait_ticket(AnyLock *lock,u32 newval)
{
u32 oldval;
// wait for our ticket to come up
// NOTE: atomic_load is [probably] overkill here
while (1) {
#if 0
oldval = atomic_load(&lock->seqacq);
#else
oldval = lock->seqacq;
#endif
if (oldval == newval)
break;
}
}
void
lock_acquire_ticket(AnyLock *lock)
{
u32 oldval;
u32 newval;
int ok;
// acquire our ticket value
// NOTE: just use a garbage value for oldval -- the exchange will
// update it with the correct/latest value -- this saves a separate
// refetch within the loop
oldval = 0;
while (1) {
#if 0
BARRIER;
oldval = lock->seqreq;
#endif
newval = oldval + 1;
ok = atomic_compare_exchange_strong(&lock->seqreq,&oldval,newval);
if (ok)
break;
}
lock_wait_ticket(lock,newval);
}
void
lock_release_ticket(AnyLock *lock)
{
// NOTE: atomic_fetch_add is [probably] overkill, but leave it for now
#if 1
atomic_fetch_add(&lock->seqacq,1);
#else
lock->seqacq += 1;
#endif
}
void
lock_acquire_mutex(AnyLock *lock)
{
pthread_mutex_lock(&lock->mutex);
}
void
lock_release_mutex(AnyLock *lock)
{
pthread_mutex_unlock(&lock->mutex);
}
void
lock_init(AnyLock *lock)
{
switch (opt_L) {
case 1:
lock->seqreq = 0;
lock->seqacq = 1;
lock_acquire = lock_acquire_ticket;
lock_release = lock_release_ticket;
break;
default:
pthread_mutex_init(&lock->mutex,NULL);
lock_acquire = lock_acquire_mutex;
lock_release = lock_release_mutex;
break;
}
}
void
startQueue(TQueue *queue)
{
queue->head = NULL;
queue->tail = NULL;
}
int
empty(TQueue *queue)
{
return (queue->head == NULL);
}
// cellnew_pool -- allocate a queue entry
Cell *
cellnew_pool(void)
{
int cnt;
Cell *p;
Cell *pool;
while (1) {
// try for quick allocation
p = cellpool;
// bug out if we got it
if (p != NULL) {
cellpool = p->next;
break;
}
// go to the heap to replenish the pool
cnt = 1000;
p = xalloc(cnt,sizeof(Cell));
// link up the entries
pool = NULL;
for (; cnt > 0; --cnt, ++p) {
p->next = pool;
pool = p;
}
// put this "online"
cellpool = pool;
}
return p;
}
// cellfree_pool -- release a queue entry
void
cellfree_pool(Cell *p)
{
p->next = cellpool;
cellpool = p;
}
// cellnew_std -- allocate a queue entry
Cell *
cellnew_std(void)
{
Cell *p;
p = xalloc(1,sizeof(Cell));
return p;
}
// cellfree_std -- release a queue entry
void
cellfree_std(Cell *p)
{
free(p);
}
void
enqueue(TQueue *queue, Task *t)
{
Cell *p;
lock_acquire(&task_mutex);
p = cellnew();
p->next = NULL;
p->t = t;
if (queue->tail == NULL) {
queue->tail = p;
queue->head = p;
}
else {
queue->tail->next = p;
queue->tail = p;
}
lock_release(&task_mutex);
}
Task *
dequeue(TQueue *queue)
{
Task *t;
lock_acquire(&task_mutex);
if (empty(queue))
t = NULL;
else {
Cell *p = queue->head;
if (p == queue->tail)
queue->tail = NULL;
queue->head = p->next;
t = p->t;
cellfree(p);
}
lock_release(&task_mutex);
return t;
}
void *
work(void *arg)
{
Thread *tskcur = arg;
TQueue *queue = tskcur->queue;
Task *t;
Log *log;
long cnt;
int tprev;
byte *bitv;
double tvbeg;
double tvbef;
double tvaft;
log = tskcur->log;
bitv = tskcur->bitv;
tvbeg = tvgetf();
tprev = 0;
while (1) {
tvbef = tvgetf();
t = dequeue(queue);
tvaft = tvgetf();
if (t == NULL)
break;
// abort if we get a double entry
if (btvset(bitv,t->number))
sysfault("work: duplicate\n");
if (opt_p) {
printf("[%.9f/%.9f %5.5d] %d [%d]\n",
tvbef,tvaft - tvbef,tskcur->xid,t->number,t->number - tprev);
tprev = t->number;
continue;
}
log->tvbef = tvbef;
log->tvaft = tvaft;
log->number = t->number;
++log;
}
if (! opt_p) {
tvaft = tvgetf();
cnt = log - tskcur->log;
log = tskcur->log;
lock_acquire(&print_mutex);
printf("\n");
printf("THREAD=%5.5d START=%.9f STOP=%.9f ELAP=%.9f TOTAL=%ld\n",
tskcur->xid,tvbeg,tvaft,tvaft - tvbeg,cnt);
tprev = 0;
for (; cnt > 0; --cnt, ++log) {
printf("[%.9f/%.9f %5.5d] %d [%d]\n",
log->tvbef,log->tvaft - log->tvbef,tskcur->xid,
log->number,log->number - tprev);
tprev = log->number;
}
lock_release(&print_mutex);
}
return (void *) 0;
}
void
btvchk(Thread *tska,Thread *tskb)
{
byte *btva;
byte *btvb;
byte aval;
byte bval;
int idx;
printf("btvchk: %d ??? %d\n",tska->xid,tskb->xid);
btva = tska->bitv;
btvb = tskb->bitv;
// abort if we get overlapping entries between two threads
for (idx = 0; idx < bitvlen; ++idx) {
aval = btva[idx];
bval = btvb[idx];
if (aval & bval)
sysfault("btvchk: duplicate\n");
}
}
// For a simple test i runned this on main:
int
main(int argc,char **argv)
{
char *cp;
TQueue *queue;
Task *t;
Thread *tsk;
--argc;
++argv;
for (; argc > 0; --argc, ++argv) {
cp = *argv;
if (*cp != '-')
break;
switch (cp[1]) {
case 'p': // print immediately
opt_p = 1;
break;
case 'Q': // number of queue items
opt_Q = atoi(cp + 2);
break;
case 'T': // number of threads
opt_T = atoi(cp + 2);
break;
case 'L':
opt_L = 1;
break;
case 'M':
opt_M = 1;
break;
default:
break;
}
}
printf("p=%d -- thread log is %s\n",opt_p,opt_p ? "immediate" : "deferred");
if (opt_T == 0)
opt_T = 16;
printf("T=%d (number of threads)\n",opt_T);
if (opt_Q == 0)
opt_Q = 1000000;
printf("Q=%d (number of items to enqueue)\n",opt_Q);
printf("L=%d -- lock is %s\n",opt_L,opt_L ? "ticket" : "mutex");
printf("M=%d -- queue item allocation is %s\n",
opt_M,opt_M ? "pooled" : "malloc/free");
tvzero = tvgetf();
lock_init(&task_mutex);
lock_init(&print_mutex);
// select queue item allocation strategy
switch (opt_M) {
case 1:
cellnew = cellnew_pool;
cellfree = cellfree_pool;
break;
default:
cellnew = cellnew_std;
cellfree = cellfree_std;
break;
}
queue = xalloc(1,sizeof(TQueue));
startQueue(queue);
Thread threads[opt_T];
// get byte length of bit vectors
bitvlen = BTVLEN(opt_Q + 1);
// allocate per-thread log buffers
for (int i = 0; i < opt_T; i++) {
tsk = &threads[i];
if (! opt_p)
tsk->log = xalloc(opt_Q,sizeof(Log));
tsk->bitv = xalloc(bitvlen,sizeof(byte));
}
// allocate "work to do"
t = xalloc(opt_Q,sizeof(Task));
// add to master queue
for (int i = 0; i < opt_Q; i++) {
t[i].number = i + 1;
enqueue(queue, &t[i]);
}
// fire up the threads
for (int i = 0; i < opt_T; i++) {
tsk = &threads[i];
tsk->xid = i + 1;
tsk->queue = queue;
pthread_create(&tsk->tid, NULL, work, tsk);
}
// wait for threads to complete
for (int i = 0; i < opt_T; i++) {
tsk = &threads[i];
pthread_join(tsk->tid, NULL);
}
// wait for threads to complete
for (int i = 0; i < opt_T; i++) {
for (int j = i + 1; j < opt_T; j++)
btvchk(&threads[i],&threads[j]);
}
printf("TOTAL: %.9f\n",tvgetf());
free(t);
return 0;
}