Question

有人可以帮助解决下面的代码。我不认为我的线程工作者正在看到信号。我希望看到STEQUESIZE 1打印，但它永远不会发生。任何帮助将不胜感激。

输出是：

线程已启动互斥锁定完成

#include <stdlib.h>
#include <stdio.h>
#include <signal.h>
#include <unistd.h>
#include <string.h>
#include <getopt.h>
#include <pthread.h>

#include <errno.h>

//  GLOBALS
pthread_mutex_t mutexqueue;
pthread_cond_t condworker;
pthread_t *threadIDs;
int STEQUESIZE=0;

void* worker(void *arg) {
 fprintf(stdout,"Thread Started\n");
 fflush(stdout);

  while (1) {
  pthread_mutex_lock(&mutexqueue);
    fprintf(stdout,"mutex locked\n");  fflush(stdout);
    while(STEQUESIZE == 0);
     pthread_cond_wait(&condworker, &mutexqueue);
    fprintf(stdout,"STEQUESIZE  %d\n",STEQUESIZE);
    fflush(stdout);
    STEQUESIZE--;

  pthread_mutex_unlock(&mutexqueue);
  }
}

int main(int argc, char **argv) {
  int nthreads = 1;

  STEQUESIZE=0;
  pthread_mutex_t mutexqueue = PTHREAD_MUTEX_INITIALIZER;
  pthread_cond_t condworker = PTHREAD_COND_INITIALIZER;

  threadIDs = (pthread_t *)malloc(sizeof(pthread_t)*nthreads);
  int j;
  for(j=0; j< nthreads;j++)
     pthread_create(&threadIDs[j], NULL, worker, NULL);

  sleep(2);

    pthread_mutex_lock(&mutexqueue);
      STEQUESIZE=1;
      pthread_cond_signal(&condworker);
    pthread_mutex_unlock(&mutexqueue);

  printf("done\n");
  while(1);
  return 0;
}

Answer 1

感谢评论回答了我的问题。我的全局变量和局部变量名称相同。

//  GLOBALS
pthread_mutex_t mutexqueue;
pthread_cond_t condworker;
pthread_t *threadIDs;
int STEQUESIZE=0;

然后

int main(int argc, char **argv) {
  int nthreads = 1;

  STEQUESIZE=0;
  pthread_mutex_t mutexqueue = PTHREAD_MUTEX_INITIALIZER;
  pthread_cond_t condworker = PTHREAD_COND_INITIALIZER;

Pthread条件信号与工人线程

1 个答案: