member of a structure pointer is also a pointer. what should temp->data represent.Question is more clear in code

Question

In first print temp->data is address whereas in later print temp->data becomes value and *temp->data prints nothing

I have tried assining value like

int n = 5;
temp->data = &n;

then *temp->data prints value 5

structure is
struct Node{
    int *data;
    struct Node *next;
};
/***part of code ***/
struct Node *temp = NULL;
    temp = head;
    while(count != l){
         count++;
         printf("%x   %x  %x ",temp,temp->data,&(temp->data));
         printf("enter %d element\n",count);
            scanf("%d",&(temp->data));
             printf("%d\n",temp->data);

Answer 1

int *data; creates an unintialized pointer and must point to valid memory before it can be used. (failure to do so leads to Undefined Behavior, and likely a SegFault)

For example:

temp->data = malloc (sizeof *temp->data);
if (temp->data == NULL) {
    perror ("malloc-temp->data");
    /* handle error, e.g. return/exit */
}

Now you can assign an integer value to temp->data. (don't forget to free the memory you allocate when it is no longer needed)

Presuming you have done that (in some other code you didn't post), the following is wrong:

 scanf("%d",&(temp->data));

temp->data is already a pointer, no '&' is needed before it.

(you must ALWAYS validate the scanf return before you consider your input valid)

&temp->data creates a pointer-to-pointer-to int (e.g. the address of the pointer to data). Printing pointer values with %x leads to undefined behavior unless sizeof (a_pointer) and sizeof (unsigned) are the same on your system (not for x86_64). Use %p to print addresses.

Address those issues and let me know if you have further problems.