C pthread和struct问题

Question

此代码的基本功能是获取计数器和线程数，创建计数器然后创建线程，然后获取每个线程中的指令数（指令格式[counter] [work-function] [重复]）

/* ============================================================================ 
* File-global variables 
* ========================================================================== */ 
static int ncounters = 0; 
static struct counter *counters = NULL; 

static int nthreads = 0; 
static int *ninstructions = NULL; 
static struct instruction **instructions = NULL; 

struct counter { 
    long long counter;   /* to store counter */ 
}; 

/* counter value */ 
struct instruction { 
    struct counter *counter;  /* pointer to counter */ 
    int repetitions;    /* number of repetitions */ 
    void (*work_fn)(long long *); /* function pointer to work function */ 
}; 

/* ============================================================================ 
* Thread function 
* ========================================================================== */ 
static void * 
worker_thread(void *arg) { 
    (void)arg; 
     int Tcounter; 
     int Trepetition; 
     char Tfuntion; 

     scanf(" %d %c %d", &Tcounter, &Tfuntion, &Trepetition); 

我如何实际存储这三个变量使用struct指令**说明？

return NULL; 
} 


/* ============================================================================ 
* Main function 
* ========================================================================== */ 
int 
main(void) { 

     scanf(" %d", &ncounters); 

     int i; 

     if(counters = malloc(sizeof(struct counter)*ncounters)){ 
      for(i=0; i < ncounters ;i++){ 
      counters[i].counter = 0; 
      } 

      for(i=0; i < ncounters ;i++){ 
      printf("%lld\n", counters[i].counter); 
      } 
     } 

     scanf(" %d", &nthreads); 
     pthread_t threads[nthreads]; 

     int ninst; 

     for(i=0; i < nthreads ;i++){ 
      scanf(" %d", &ninst); 
      ninstructions = ninst; 
      for(i=0; i < ninstructions ;i++){ 
      pthread_create(&threads[i], NULL, worker_thread, NULL); 
      } 
     } 

     free(counters); 
    return 0; 
} 

pthread_create函数是否正确？

int 
main(void) { 

     scanf(" %d", &ncounters); 

     int i; 

     if(counters = malloc(sizeof(struct counter)*ncounters)){ 
      for(i=0; i < ncounters ;i++){ 
      counters[i].counter = 0; 
      } 
     } 

     scanf(" %d", &nthreads); 

     pthread_t threads[nthreads]; 

     if(ninstructions = (int*)malloc(nthreads*sizeof(int)){ 
      for(i=0; i < nthreads ;i++){ 
      scanf(" %d", &ninstructions[i]); 
      int j; 
      for(j=0; i < ninstructions[i] ;j++){ 
       pthread_create(&threads[j], NULL, worker_thread, NULL); 
      } 
      } 
     } 


     free(ninstructions); 
     free(counters); 
    return 0; 
} 

我也试图在ninstruction变成数组，如果它的正确...

Answer 1

OK，下一个尝试。这一次在伪代码，因为这味道像功课...

struct instruction 
{ 
    long long counter; 
    int repetitions 
}; 


main() 
{ 
    ask #threads 

    pthread_t threads[nthreads]; 
    struct instruction intructions[nthreads]; 

    while(i < nthreads) 
    { 
    pthread_create(&threads[i], NULL, threadMain, &instructions[i]); 
    } 

    i = 0 
    while (i < nthreads) 
    { 
    //now wait until all threads have finished 
    pthread_join(threads[i], NULL); 
    } 
} 

    void threadMain(void* arg) 
    { 
    struct instruction* pInstruction = (struct instruction*)arg; 
    char cFunctionCode; 

    enter protected section, otherwise all threads will ask at the same time 

    scanf(" %d %c %d", &pInstruction->counter, &cFunctionCode, &pInstruction->repetitions 

    leave protected section here 

    do youtr computes here 

    return; 
) 

对于受保护的部分，查找互斥体或信号量。

Answer 2

确定。我想你要做到以下几点：

... 
//create the instructions, which is missing. 
// be carefull that instruction.counter is not allocated!!! 

struct instruction instructions[ncounters]; 
... 
pthread_create(&threads[i], NULL, worker_thread, &instructions[i]); 
... 

而且在工作线程

worker_thread(void* arg) 
{ 
    struct instruction* pInstruction = (struct instruction*)arg; 

    pInstruction->repetions = 42; 
... 

我希望这是你想要的东西......

马里奥