Delete employee from doubly linked list

Assume each employee node is:

struct emp {
    struct emp *prev;
    char id[10];           /* e.g. "165-64" */
    char name[30];
    float salary;
    struct emp *next;
};
struct emp *start;         /* first node */

Idea

Search the list for the node whose ID is "165-64"; then make its previous node and next node point to each other, and free it. There are three cases: first node, middle node, last node.

Illustration

Before:
start -> [NULL|101-12|*] <-> [*|165-64|*] <-> [*|201-07|NULL]
                               ^ p

After:
start -> [NULL|101-12|*] <-> [*|201-07|NULL]

The first node's next is set to the third node, and the third node's prev is set to the first node. Node "165-64" is freed.

Pseudo code

DELETE_EMP(start, "165-64")
1. if start == NULL then print "empty list"; return
2. p = start
3. while p != NULL and p->id != "165-64"
       p = p->next
4. if p == NULL then print "employee not found"; return
5. if p->prev == NULL            // deleting first node
       start = p->next
   else
       p->prev->next = p->next
6. if p->next != NULL            // not the last node
       p->next->prev = p->prev
7. free(p)

C code

void delete_emp(struct emp **start, const char *id) {
    struct emp *p = *start;
    while (p != NULL && strcmp(p->id, id) != 0)
        p = p->next;
    if (p == NULL) { printf("Not found\n"); return; }
    if (p->prev == NULL) *start = p->next;      /* first node */
    else p->prev->next = p->next;
    if (p->next != NULL) p->next->prev = p->prev;
    free(p);
}
/* call: delete_emp(&start, "165-64"); */

Cases: if it is the first node, start moves to the second node; if it is the only node, start becomes p->next, which is NULL, and step 6 is skipped; if it is the last node, step 6 is skipped. Time is O(n) for the search and O(1) for the removal.