Hey! wait a minute.....forget my previous reply.
where are you registering your open fxn. in the fops struct???
Yeah !! this is the only, way i know....
static struct file_operations my_fops = {
.read = my_read,
.write = my_write,
.open = my_open,
.release = my_release
};
and this is the snippet of the code.... where i am doing this
my_cdev.ops =&my_fops;
rv = cdev_add(&my_cdev,devno,1);
(I'm assuming you have shown us your complete code)
Yes...
Your open is not going to get called automagically! :)
this is my_open implementation
static int my_open(struct inode *imem, struct file *fmem)
{
printk("<1> INSIDE my_open ..................\n");
return 1;
}
Regards,
Vishal.
Vishal.
