On strings and memory allocation:
A string in C is just a sequence of chars, so you can use char * or a char array wherever you want to use a string data type:
typedef struct {
int number;
char *name;
char *address;
char *birthdate;
char gender;
} patient;
Then you need to allocate memory for the structure itself, and for each of the strings:
patient *createPatient(int number, char *name,
char *addr, char *bd, char sex) {
// Allocate memory for the pointers themselves and other elements
// in the struct.
patient *p = malloc(sizeof(struct patient));
p->number = number; // Scalars (int, char, etc) can simply be copied
// Must allocate memory for contents of pointers. Here, strdup()
// creates a new copy of name. Another option:
// p->name = malloc(strlen(name)+1);
// strcpy(p->name, name);
p->name = strdup(name);
p->address = strdup(addr);
p->birthdate = strdup(bd);
p->gender = sex;
return p;
}
If you'll only need a few patients, you can avoid the memory management at the expense of allocating more memory than you really need:
typedef struct {
int number;
char name[50]; // Declaring an array will allocate the specified
char address[200]; // amount of memory when the struct is created,
char birthdate[50]; // but pre-determines the max length and may
char gender; // allocate more than you need.
} patient;
On linked lists:
In general, the purpose of a linked list is to prove quick access to an ordered collection of elements. If your llist contains an element called num (which presumably contains the patient number), you need an additional data structure to hold the actual patients themselves, and you'll need to look up the patient number every time.
Instead, if you declare
typedef struct llist
{
patient *p;
struct llist *next;
} list;
then each element contains a direct pointer to a patient structure, and you can access the data like this:
patient *getPatient(list *patients, int num) {
list *l = patients;
while (l != NULL) {
if (l->p->num == num) {
return l->p;
}
l = l->next;
}
return NULL;
}