我想在 Linux 上使用(Unix 98 风格的)伪 tty接收(以及以后的进程)write(1)
和消息。wall(1)
我已经有以下最小实现:
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
#include <unistd.h>
#include <errno.h>
#include <fcntl.h>
#include <signal.h>
#include <utempter.h>
#define BUF_LENGTH 1024
int
main (void)
{
FILE *lf;
int masterfd, slavefd;
char *slave_name = NULL;
char buf[BUF_LENGTH];
size_t nbytes = sizeof(buf);
ssize_t bytes_read;
int exit_code = EXIT_SUCESS;
if ((masterfd = posix_openpt (O_RDWR | O_NOCTTY)) == -1
|| grantpt (masterfd) == -1
|| unlockpt (masterfd) == -1
|| (slave_name = ptsname (masterfd)) == NULL)
exit (EXIT_FAILURE);
if (!(lf = fopen("term.log","w")))
exit (EXIT_FAILURE);
addToUtmp (slave_name, NULL, masterfd);
for (;;)
{
bytes_read = read(masterfd, buf, nbytes);
if (bytes_read <= 0)
break
fwrite (buf, 1, bytes_read, lf);
}
if (bytes_read < 0)
{
fprintf (stderr, "error reading from master pty: %s\n", strerror (errno));
exit_code = EXIT_FAILURE;
}
fclose (lf);
if (slavefd >= 0)
close (slavefd);
if (masterfd >= 0)
{
removeLineFromUtmp (slave_name, masterfd);
close (masterfd);
}
exit (exit_code);
}
现在的问题是它只适用于阅读第一条消息,然后 read 给了我一个 EIO 错误。这是为什么?