On Jan 6, 2008 6:58 PM, Onkar <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> i am implementing scull (character device driver in LDD : chapter 3 ) , in
> that I found ::
>
> Please refer to Linux Device Drivers Rubunu et al , 3/e.pg 57
>
> ssize_t scull_read(struct file *filp, char __user *buf, size_t count,loff_t
> *f_pos)
>
> Scull_Dev *scull_follow(Scull_Dev *dev, int n)
> {
> while (n--) {
> if (!dev->next) {
> dev->next = kmalloc(sizeof(Scull_Dev), GFP_KERNEL);
> <<<============== why memory allocations in read operations??
> memset(dev->next, 0, sizeof(Scull_Dev));
> }
> dev = dev->next;
> continue;
> }
> return dev;
> }
>
> IMHO this implementation is WRONG ?? Correct me if I am wrong.
>
> other implementation is :
>
> /*
> scull_follow
> Take a walk in the linked list until node 'n' has been reached.
> */
> struct scull_qset* scull_follow(struct scull_dev *dev, int n){
> struct scull_qset *qs = dev->data;
>
> // Allocate first node if needed
Only if qs == NULL then allocate memory. (In case of write)
else skip this statement block altogether. For read and if qs has
some persistent data already, this statement block will be skipped.
> if(!qs) {
> // Reserve memory
> qs = dev->data = kmalloc(sizeof(struct scull_qset), GFP_KERNEL);
> // Failure :(
> if(qs == NULL){
> return NULL;
> }
> // Zero out the region
> memset(qs, 0, sizeof(struct scull_qset));
> }
Same below.
>
>
> while(n--){
> // If the next node does not exist, allocate it.
> if(!qs->next){
> qs->next = kmalloc(sizeof(struct scull_qset), GFP_KERNEL);
> if(qs->next== NULL){
> return NULL;
> }
> memset(qs->next, 0, sizeof(struct scull_qset));
> }
> qs = qs->next;
> continue;
> }
> return qs;
> }
>
> But here too i am not convinced with memory allocation ?? Why memory
> allocations in read when we only
> want to reach the first partially filled node ??
>
>
Thanks.
