#include <stdio.h>
#include <semaphore.h>
#include <pthread.h>
#include <fcntl.h>

pthread_mutex_t mutex=PTHREAD_MUTEX_INITIALIZER;
sem_t* t=NULL;

void* thread(void* arg)
{
    //sleep(10);
    int status=pthread_mutex_lock(&mutex);
    if(status)
    {
        printf("can't lock mutex\n");
        exit(1);
    }
   
    sem_wait(t);
    int value;
    printf("pthread id is %d get resource\n",pthread_self());
    sem_getvalue(t,&value);
    printf("now the semaphore value is %d\n",value);
   
    status=pthread_mutex_unlock(&mutex);
    if(status)
    {
        printf("can't not unlock mutex\n");
        exit(1);
    }
}

int main()
{
    pthread_t pid;
    int status=pthread_mutex_lock(&mutex);
    if(status)
    {
        printf("can't lock mutex\n");
        exit(1);
    }
    sem_init(t,0,3);
    status=pthread_mutex_unlock(&mutex);
    if(status)
    {
        printf("can't unlock mutex\n");
        exit(1);
    }

    int i=0;
    while(i++<3)
        if((pthread_create(&pid,NULL,thread,NULL))!=0)
        {
            printf("pthread can't create.\n");
            exit(0);
        }

    pthread_join(pid,NULL);
    sem_destroy(t);
    return 0;
}   

在Windows下跑没有segmentation fault，，，
在Linux下就出现了。
求解

已解决
定义时应该写成这样:
sem_t t;//写成这样，sem_t *t;这是一个指针，没有初始化，没有给他非配内存它这时候这时候只是一个也指针， 自然会出错。