我已经编写了两个程序：第一个程序，即“写程序”，创建一个FIFO并将数据写入其中。第二个“读取器”在后台运行，并在FIFO中查找数据。一旦有数据，读取器就会将其读出。
如果我开始例如两个写入器和两个读取器，它们都可以在同一个FIFO中写入/读取。如何限制第三和第四读取器/写入器使用FIFO，并只允许一个写入器和一个读取器使用FIFO？
我的代码：
FIFO写入器：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
#define CHMOD 0777
int main(int argc, char **argv)
{
char outbuf[BUFFERSIZE]; // outbuffer
int fifo, j, anzahl;
// fifo - pipe file deskriptor, j - counter, anzahl - Parameter.
if(argc!=2) // Check if parameter is ok
{
printf("Ungültiger Parameter! Bsp.: ./fifow 10\n");
return 1;
}
anzahl=atoi(argv[1]); // convert paramter to integer
mkfifo("namedpipe4", CHMOD); // make FIFO "namedpipe4"
fifo = open("namedpipe4",O_WRONLY); // open FIFO
//
for(j=0;j<anzahl;j++)
{
printf("Writer PID: %d writes record nr. %6d\n", getpid(), j+1);
sprintf(outbuf, "Writer PID: %d writes record nr. %6d\n", getpid(), j+1);
write(fifo, outbuf, BUFFERSIZE);
remove("namedpipe4"); // removing the fifo
sleep(1); // Wait 1 sec
}
close(fifo); //
exit(0);
}
FIFO读取器：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
int main(void)
{
char inbuf[BUFFERSIZE]; // inbuffer
int fifo, var;
printf("\n Waiting for a Pipe....\n");
while((fifo = open("namedpipe4",O_RDONLY)) == -1) // while "there is no such pipe"
{
remove("namedpipe4");
sleep(1);
}
while((var = read(fifo, inbuf, BUFFERSIZE)) > 0) // while "i can read"
{
printf("Reader PID: %d reads record: %s\n", getpid(), inbuf);
sleep(1);
}
close(fifo); //
printf("\n EOF..\n");
exit(0);
}
给出the code you posted in a separate answer，这是修改的版本，可以解决您遇到的问题。详见注释，但总而言之：
mkfifo的返回值，以检查是否有其他作者已创建管道。flock获得了管道的排他性咨询锁定，以避免在第二个读者在第一个读者删除管道之前就已经打开管道的竞争情况。作家：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/stat.h> /* needed for mkfifo */
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
#define CHMOD 0777
int
main (int argc, char **argv)
{
char outbuf[BUFFERSIZE];
int fifo, j, anzahl;
if (argc != 2)
{
printf("Ungültiger Parameter! Bsp.: ./fifow 10\n");
return 1;
}
anzahl=atoi(argv[1]);
/* mkfifo fails if the file already exists, which means there's a
* writer waiting for a reader. This assures that only one writer
* will write to the pipe, since it only opens the pipe if it was
* the one who created it.
*/
if (mkfifo("namedpipe4", CHMOD) == -1)
{
printf("namedpipe4 already exists\n");
return 1;
}
fifo = open("namedpipe4", O_WRONLY);
for (j = 0; j < anzahl; j++)
{
printf("Writer PID: %d writes record nr. %6d\n", getpid(), j + 1);
sprintf(outbuf, "Writer PID: %d writes record nr. %6d\n", getpid(), j + 1);
write(fifo, outbuf, BUFFERSIZE);
remove("namedpipe4");
sleep(1);
}
close(fifo);
exit(0);
}
读者：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/file.h> /* for flock */
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
int
main (int argc, char **argv)
{
char inbuf[BUFFERSIZE];
int fifo, var;
printf("\n Waiting for a Pipe....\n");
/* There are *two* ways the open can fail: the pipe doesn't exist
* yet, *or* it succeeded, but a different writer already opened
* it but didn't yet remove it.
*/
while (1)
{
while ((fifo = open("namedpipe4", O_RDONLY)) == -1)
{
/* Since you didn't specify O_CREAT in the call to open, there
* is no way that namedpipe4 would have been created by the
* reader. If there *is* now a namedpipe4, a remove here
* would delete the one the writer created!
*/
sleep(1);
}
/* Get an exclusive lock on the file, failing if we can't get
* it immediately. Only one reader will succeed.
*/
if (flock (fifo, LOCK_EX | LOCK_NB) == 0)
break;
/* We lost the race to another reader. Give up and wait for
* the next writer.
*/
close (fifo);
}
/* We are definitely the only reader.
*/
/* *Here* we delete the pipe, now that we've locked it and thus
* know that we "own" the pipe. If we delete before locking,
* there's a race where after we opened the pipe, a different
* reader also opened, deleted, and locked the file, and a new
* writer created a new pipe; in that case, we'd be deleting the
* wrong pipe.
*/
remove("namedpipe4");
while ((var = read(fifo, inbuf, BUFFERSIZE)) > 0)
{
printf("Reader PID: %d reads record: %s\n", getpid(), inbuf);
/* No need to sleep; we'll consume input as it becomes
* available.
*/
}
close(fifo);
printf("\n EOF..\n");
exit(0);
}
使用pipe(2)创建FIFO，并且仅在从父进程派生它们的时候，才将FIFO两端的文件描述符提供给适当的进程。 （或者，让读者调用pipe(2)并派发编写者，反之亦然。）由于FIFO从未存在于文件系统上，因此任何其他进程都无法访问它。
如果必须使用命名的FIFO，请在读写器打开FIFO后将其删除。只要读取器和写入器将其打开，底层的FIFO将仍然存在，但是没有新的进程将能够打开它。但是，在竞争状态下，第二个读取器或写入器可以在删除FIFO之前打开它。
FIFO写入器：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
#define CHMOD 0777
int main(int argc, char **argv)
{
char outbuf[BUFFERSIZE]; // outbuffer
int fifo, j, anzahl;
// fifo - pipe file deskriptor, j - counter, anzahl - Parameter.
if(argc!=2) // Check if parameter is ok
{
printf("Ungültiger Parameter! Bsp.: ./fifow 10\n");
return 1;
}
anzahl=atoi(argv[1]); // convert paramter to integer
mkfifo("namedpipe4", CHMOD); // make FIFO "namedpipe4"
fifo = open("namedpipe4",O_WRONLY); // open FIFO
//
for(j=0;j<anzahl;j++)
{
printf("Writer PID: %d writes record nr. %6d\n", getpid(), j+1);
sprintf(outbuf, "Writer PID: %d writes record nr. %6d\n", getpid(), j+1);
write(fifo, outbuf, BUFFERSIZE);
remove("namedpipe4"); // removing the fifo
sleep(1); // Wait 1 sec
}
close(fifo); //
exit(0);
}
FIFO读取器：
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <fcntl.h>
#define BUFFERSIZE 50
int main(void)
{
char inbuf[BUFFERSIZE]; // inbuffer
int fifo, var;
printf("\n Waiting for a Pipe....\n");
while((fifo = open("namedpipe4",O_RDONLY)) == -1) // while "there is no such pipe"
{
remove("namedpipe4");
sleep(1);
}
while((var = read(fifo, inbuf, BUFFERSIZE)) > 0) // while "i can read"
{
printf("Reader PID: %d reads record: %s\n", getpid(), inbuf);
sleep(1);
}
close(fifo); //
printf("\n EOF..\n");
exit(0);
}