【发布时间】:2014-08-14 14:21:57
【问题描述】:
我有一个 C 程序,父进程创建一个读取器线程,然后派生一个子进程,子进程创建多个写入器线程。
写入线程正确地将元素插入共享缓冲区,但读取线程不做任何事情!当我将所有读取器和写入器线程放在一个进程中时,程序正常工作并且读取器从缓冲区中读取元素!读取器和写入器都使用sem_t 作为信号量,使用sem_wait 和sem_post 管理对缓冲区的访问。
这是我的程序伪代码:
int main()
{
initialize semaphores
create reader_thread(reader code)
fork child process(server)
}
int server()
{
create writer threads(writer code)
}
这是缓冲区结构:
typedef struct
{
Req_info reqinfo[BUFFER_SIZE];
char chunk[BUFFER_SIZE][MAX_CHUNK_SIZE];
uint64_t chunk_size[BUFFER_SIZE];
int Isnewfile[BUFFER_SIZE]; //1 means new file
int Islastchunk[BUFFER_SIZE]; //1 means last chunk
int ID[BUFFER_SIZE];
int head;
int tail;
int numofelement;
#ifdef SEM_V
sem_t full;
sem_t empty;
sem_t mutex;
#else
pthread_mutex_t mutex;
pthread_cond_t cond_read;
pthread_cond_t cond_write;
#endif
} Buffer;
Buffer *buffer_ptr;
void writer()
{
#ifdef SEM_V
sem_wait(&buffer_ptr->empty);
sem_wait(&buffer_ptr->mutex);
#else
pthread_mutex_lock(&buffer_ptr->mutex);
if((buffer_ptr->tail + 1) % BUFFER_SIZE == buffer_ptr->head)
pthread_cond_wait( &buffer_ptr->cond_write, &buffer_ptr->mutex );
pthread_mutex_unlock(&buffer_ptr->mutex);
pthread_mutex_lock(&buffer_ptr->mutex);
#endif
if ((buffer_ptr->tail + 1) % BUFFER_SIZE != buffer_ptr->head)
{
memmove(buffer_ptr->chunk[buffer_ptr->tail], chunk, chunk_size); //Write chunk into buffer
buffer_ptr->chunk[buffer_ptr->tail][chunk_size] = '\0';
buffer_ptr->chunk_size[buffer_ptr->tail] = chunk_size; //Write chunk size into buffer
buffer_ptr->Isnewfile[buffer_ptr->tail] = Isnewfile;
buffer_ptr->Islastchunk[buffer_ptr->tail] = Islastchunk;
buffer_ptr->reqinfo[buffer_ptr->tail] = reqinfo;
buffer_ptr->ID[buffer_ptr->tail] = ID;
buffer_ptr->tail = (buffer_ptr->tail + 1) % BUFFER_SIZE;
}
#ifdef SEM_V
sem_post(&buffer_ptr->mutex);
sem_post(&buffer_ptr->full);
#else
pthread_cond_signal(&buffer_ptr->cond_write);
pthread_mutex_unlock(&buffer_ptr->mutex);
#endif
}
void reader()
{
#ifdef SEM_V
sem_wait(&buffer_ptr->full);
#endif
if (buffer_ptr->tail != buffer_ptr->head)
{
if(!first){
gettimeofday(&ts, NULL);
first = 1;
}
chunksize = buffer_ptr->chunk_size[buffer_ptr->head]; //Read chunk size from buffer
memmove(chunk, buffer_ptr->chunk[buffer_ptr->head], chunksize); //Read chunk from buffer
chunk[chunksize] = '\0';
Isnewfile = buffer_ptr->Isnewfile[buffer_ptr->head];
Islastchunk = buffer_ptr->Islastchunk[buffer_ptr->head];
reqinfo = buffer_ptr->reqinfo[buffer_ptr->head];
ID = buffer_ptr->ID[buffer_ptr->head];
buffer_ptr->head = (buffer_ptr->head + 1) % BUFFER_SIZE;
}
else{
#ifdef SEM_V
sem_post(&buffer_ptr->empty);
#endif
continue;
}
#ifdef SEM_V
sem_post(&buffer_ptr->empty);
#endif
}
【问题讨论】:
-
你在读/写什么?如果它只是一个普通的 C 缓冲区,它将无法工作。您需要使用共享内存。
-
至少显示一些实际代码怎么样?
-
@Drew McGowen,每个写入线程逐块读取图像文件并将其写入共享缓冲区。共享缓冲区是一个包含一些字段数组的结构。读取线程从缓冲区中读取元素并将它们写入一个存储文件。
-
它实际上是共享内存吗?就像@ScottHunter 所说,你应该发布 actual 代码;你的伪代码描述不够。
-
这是你的问题 - 普通缓冲区不会在
fork之间隐式共享。
标签: c multithreading semaphore