How to work with string fields in a C struct?

Question

I'm having trouble making a database based on a singly-linked list in C, not because of the linked list concept but rather the string fields in the struct themselves.

This is an assignment in C and as far as I know (I'm a newbie), C doesn't recognize 'string' as a data type.

This is what my struct code looks like:

typedef struct 
{
  int number;
  string name;
  string address;
  string birthdate;
  char gender;
} patient;

typedef struct llist
{
  patient num;
  struct llist *next;
} list;

I was thinking of making a struct for the strings themselves so that I can use them in the struct, like this:

typedef struct string 
{ 
  char *text;
} *string;

Then I will malloc() each one of them when it is required to make new data of the string type (array of char).

typedef struct string
{
  char *text;
} *string;

int main()
{
    int length = 50;
    string s = (string) malloc(sizeof string);
    s->text = (char *) malloc(len * sizeof char);
    strcpy(s->text, patient.name->text);
}

Can someone help me figure this out?
Thank you.

Answer 1

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;
}

Answer 2

I think this solution uses less code and is easy to understand even for newbie.

For string field in struct, you can use pointer and reassigning the string to that pointer will be straightforward and simpler.

Define definition of struct:

typedef struct {
  int number;
  char *name;
  char *address;
  char *birthdate;
  char gender;
} Patient;

Initialize variable with type of that struct:

Patient patient;
patient.number = 12345;
patient.address = "123/123 some road Rd.";
patient.birthdate = "2020/12/12";
patient.gender = 'M';

It is that simple. Hope this answer helps many developers.

Answer 3

While Richard's is what you want if you do want to go with a typedef, I'd suggest that it's probably not a particularly good idea in this instance, as you lose sight of it being a pointer, while not gaining anything.

If you were treating it a a counted string, or something with additional functionality, that might be different, but I'd really recommend that in this instance, you just get familiar with the 'standard' C string implementation being a 'char *'...

Answer 4

This does not work:

string s = (string)malloc(sizeof string); 

string refers to a pointer, you need the size of the structure itself:

string s = malloc(sizeof (*string)); 

Note the lack of cast as well (conversion from void* (malloc's return type) is implicitly performed).

Also, in your main, you have a globally delcared patient, but that is uninitialized. Try:

 patient.number = 3;     
 patient.name = "John";     
 patient.address = "Baker street";     
 patient.birthdate = "4/15/2012";     
 patient.gender = 'M';     

before you read-access any of its members

Also, strcpy is inherently unsafe as it does not have boundary checking (will copy until the first '\0' is encountered, writing past allocated memory if the source is too long). Use strncpy instead, where you can at least specify the maximum number of characters copied -- read the documentation to ensure you pass the correct value, it is easy to make an off-by-one error.

Answer 5

You could just use an even simpler typedef:

typedef char *string;

Then, your malloc would look like a usual malloc:

string s = malloc(maxStringLength);