问个linux驱动一书一个例子的问题

[xiangye]

linux驱动一书第二版 scull例子；先定义了一个函数指针数组

struct file_operations *scull_fop_array[]={
&scull_fops, /* type 0 */
&scull_priv_fops, /* type 1 */
&scull_pipe_fops, /* type 2 */
&scull_sngl_fops, /* type 3 */
&scull_user_fops, /* type 4 */
&scull_wusr_fops /* type 5 */
};

open函数：
int scull_open(struct inode *inode, struct file *filp)
{
Scull_Dev *dev; /* device information */
int num = NUM(inode->i_rdev);
int type = TYPE(inode->i_rdev);
if (!filp->private_data && type) {
if (type > SCULL_MAX_TYPE) return -ENODEV;
filp->f_op = scull_fop_array[type];
return filp->f_op->open(inode, filp); /* dispatch to specific op
en */

}

/* type 0, check the device number (unless private_data valid) */
dev = (Scull_Dev *)filp->private_data;
if (!dev) {
if (num >= scull_nr_devs) return -ENODEV;
dev = &scull_devices[num];
filp->private_data = dev; /* for other methods */
}
MOD_INC_USE_COUNT; /* Before we maybe sleep */
/* now trim to 0 the length of the device if open was write-only */
if ( (filp->f_flags & O_ACCMODE) == O_WRONLY) {
if (down_interruptible(&dev->sem)) {
MOD_DEC_USE_COUNT;
return -ERESTARTSYS;
}
scull_trim(dev); /* ignore errors */
up(&dev->sem);
}
return 0; /* success */
}

我的问题就是：这个open函数和函数指针数组第一项指针指向的函数处理的open函
数是一样的。

struct file_operations scull_fops = {
llseek: scull_llseek,
read: scull_read,
write: scull_write,
ioctl: scull_ioctl,
open: scull_open,
release: scull_release,
};
这样在scull_open函数中 return filp->f_op->open(inode, filp);时不会递归调
用了吗

[xiangye]

在scull_open函数中 return filp->f_op->open(inode, filp)，当filp->fop=&scull_fops时不会递归调用了吗

[aria]

注意在前面：if (!filp->private_data && type) {
已经保证type不会为0，因此不会递归调用。

[slone]

if (!filp->private_data && type) {
if (type > SCULL_MAX_TYPE) return -ENODEV;
filp->f_op = scull_fop_array[type];
return filp->f_op->open(inode, filp); /* dispatch to specific open */
}
IF 条件说明 指定filp 设备节点与本OPEN 所在的设备节点是不一样的。