带工作线程的线程状态信号

Question

有人可以帮助下面的代码。我不认为信号正在被我的线程工作人员看到。我期望看到STEQUESIZE 1打印，但它从来没有发生过。任何帮助，将不胜感激。

输出为：

开始线程 互斥锁 做

#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

感谢评论回答我的问题。我有全局和局部变量命名相同。

// 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;