如何強制一個線程在另一個之前啓動？

Question

例如，我想創建5個線程並打印它們。我如何在第二個之前執行第四個？我試着用互斥體鎖定它，但我不知道如何只鎖定第二個，所以它給了我分段錯誤。

char name1[]="THREAD1"; 
char name2[]="THREAD2"; 
char name3[]="THREAD3"; 
char name4[]="THREAD4"; 
char name5[]="THREAD5"; 

pthread_mutex_t mutex; 

pthread_t t1, t2,t3,t4,t5; 
int* state[5]={0,0,0,0,0}; 

void* execThread(void* threadName) 
{ 
int i; 
int* number = (int*) malloc(sizeof(int)); 
char* tmp; 

if(*(state[4]) == 0 && (pthread_self()==t2)) 
pthread_mutex_lock (&mutex); 
else if(*(state[4]) == 0 && (pthread_self()!=t2)) 
{ 

pthread_yield(); 
printf("[Thread %s] Executing step %d\n", (char *) threadName, i); 

tmp = & ((char*)threadName)[strlen((char*)threadName)-1]; 
*number = atoi(tmp); 

return number; 
} 
else 
{ 
    pthread_mutex_unlock (&mutex); 

    pthread_yield(); 
    printf("[Thread %s] Executing step %d\n", (char *) threadName, i); 


    tmp = & ((char*)threadName)[strlen((char*)threadName)-1]; 
    *number = atoi(tmp); 

    return number; 
} 
} 


void main() 
{ 

pthread_mutex_init(&mutex, NULL); 


printf("[Thread MAIN] Starting. tid=%ld, pid=%d\n", (long int) pthread_self(), (int) getpid()); 

if (pthread_create(&t1, NULL, execThread, name1)!=0) { 
perror("Error creating a new thread"); 
exit(1); 
} 

if (pthread_create(&t2, NULL, execThread, name2)!=0) { 
perror("Error creating a new thread"); 
exit(1); 
}  

if (pthread_create(&t3, NULL, execThread, name3)!=0) { 
perror("Error creating a new thread"); 
exit(1); 
} 
if (pthread_create(&t4, NULL, execThread, name4)!=0) { 
perror("Error creating a new thread"); 
exit(1); 
} 
if (pthread_create(&t5, NULL, execThread, name5)!=0) { 
perror("Error creating a new thread"); 
exit(1); 
} 


printf("[Thread MAIN] Created 5 threads \n"); 

pthread_join(t1, (void**) &state[0]); 
pthread_join(t2, (void**) &state[1]); 
pthread_join(t3, (void**) &state[2]); 
pthread_join(t4, (void**) &state[3]); 
pthread_join(t5, (void**) &state[4]); 

printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name1, (long int) t1, *(state[0])); 
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name2, (long int) t2, *(state[1])); 
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name3, (long int) t3, *(state[2])); 
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name4, (long int) t4, *(state[3])); 
printf("[Thread MAIN] Thread %s[%ld] terminated with state %d\n", name5, (long int) t5, *(state[4])); 

printf("[Thread MAIN] Terminating. tid=%ld, pid=%d\n", (long int) pthread_self(), (int) getpid()); 


pthread_mutex_destroy(&mutex); 
//pthread_exit(NULL); 
} 

Answer 1

如果要螺紋2只螺紋4已打印它的消息後打印它的消息，則可以使用一個標誌和條件變量。請注意，您無法安全地訪問線程中的t1，t2等變量而無需進一步同步（它們正在由主線程寫入，並且您正嘗試從其他線程讀取它們）。

int t4_done = 0; 
pthread_mutex_t t4_done_mutex = PTHREAD_MUTEX_INITIALIZER; 
pthread_cond_t t4_done_cond = PTHREAD_COND_INITIALIZER; 

void* execThread(void* threadName) 
{ 
    if (threadName == name2) 
    { 
     /* thread2 waits for thread4 to set the t4_done flag */ 
     pthread_mutex_lock(&t4_done_mutex); 
     while (!t4_done) 
      pthread_cond_wait(&t4_done_cond, &t4_done_mutex); 
     pthread_mutex_unlock(&t4_done_mutex); 
    } 

    printf("[Thread %s] Executing step\n", (char *) threadName); 

    if (threadName == name4) 
    { 
     /* thread4 sets the t4_done flag and signals thread2 */ 
     pthread_mutex_lock(&t4_done_mutex); 
     t4_done = 1; 
     pthread_cond_signal(&t4_done_cond); 
     pthread_mutex_unlock(&t4_done_mutex); 
    } 

    return NULL; 
}