求解linux cat命令的实现机制！cat设备节点文件，没任何输出（之前已写入）
pipe设备：实现一个循环缓冲区阻塞读写pipe0为设备节点
应用程序：
int main(int argc, char **argv)
{
int fd;
int err;
char wb = "today is good day!";
char wb1 = "what's the problem";
char rb ;
fd = open("/dev/pipe0", O_RDWR);//读写打开设备
if (fd < 0)
{
printf("can't open!\n");
}
printf("fd is : %d\n ",fd);
printf("%s\n", wb);
printf("%s\n", wb1);
write(fd, &wb, 20);//写二次
write(fd, &wb1, 20);
err=read(fd, &rb, 20);//读数据
printf("err is :%d\n", err);
printf("%s\n", rb);//打印所读到数据
return 0;
}
应用程序Linux下运行结果：
root@book-desktop:/work/pipe# ./pipetest
fd is : 3
today is good day!
what's the problem
err is :20
today is good day!
root@book-desktop:/work/pipe#
root@book-desktop:/work/pipe#
root@book-desktop:/work/pipe# cat /dev/pipe0
^C
root@book-desktop:/work/pipe#
cat pipe0怎么什么都没有，是无数据还是阻塞，为什么？求大神讲解下cat的机制啊? 应该跟cat没啥关系，cat的读写肯定都是block io。
可以strace确认一下，是block在open上，还是block在read上。
可以找一下fifo/pipe的实现，依稀记得，它对应的kernel buffer不是固定不变的，好像跟reader/writer的引用计数有关系。
如果你的程序不退出，在return 0之前加个pause，也许就能被外面的cat读到了。【不确定，自己做实验试试吧】static int fifo_open(struct inode *inode, struct file *filp)
{
pipe = inode->i_pipe;
if (!pipe) {
pipe = alloc_pipe_info(inode);
inode->i_pipe = pipe;
}
switch (filp->f_mode) {
case 3:
/*
*O_RDWR
*POSIX.1 leaves this case "undefined" when O_NONBLOCK is set.
*This implementation will NEVER block on a O_RDWR open, since
*the process can at least talk to itself.
*/
filp->f_op = &rdwr_fifo_fops;
pipe->readers++;
pipe->writers++;
}
const struct file_operations rdwr_fifo_fops = {
.release = pipe_rdwr_release,
};
static int pipe_rdwr_release(struct inode *inode, struct file *filp)
{
int decr, decw;
decr = (filp->f_mode & FMODE_READ) != 0;
decw = (filp->f_mode & FMODE_WRITE) != 0;
return pipe_release(inode, decr, decw);
}
static int pipe_release(struct inode *inode, int decr, int decw)
{
pipe->readers -= decr;
pipe->writers -= decw;
if (!pipe->readers && !pipe->writers) {
free_pipe_info(inode);
}
}
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