Thread: Implementing a medical office. – Linked List

So far i got new patient and show_queue (working).

I got a little doubt

When i make malloc should i put the patient i just create to NULL or that don´t matter ?

Code:

    // 1º Get Memory from the new patient, and putting the block to NULL
    if ((new_patient = malloc(sizeof (node))) == NULL) {
        printf("Something went wrong -> %s\n", strerror(errno));
        return list;
    }
    new_patient->next = NULL; // Put the NULL the patient i just create

Full Functions

Code:

node * new_patient(node * list) {


    node * new_patient;
    node * aux;


    // 1º Get Memory from the new patient, and putting the block to NULL
    if ((new_patient = malloc(sizeof (node))) == NULL) {
        printf("Something went wrong -> %s\n", strerror(errno));
        return list;
    }
    new_patient->next = NULL; // Put the NULL the patient i just create


    // 2º Getting the input from the user
    puts("What is the name of the new patient ?");
    scanf(" %[^\n]", new_patient->name);


    // 3º Insert on the list
    if (list == NULL) {
        list = new_patient;
    } else {
        aux = list;


        // Going to run the list until i reach the end.
        while (aux->next != NULL) {
            aux = aux->next;
        }


        // Put the last one pointing to the new block
        aux->next = new_patient;
    }


    return list;
}


void show_queue(node * list) {
    
    if (list == NULL) {
        printf("The list is empry\n");
    } else {
        while (list != NULL) {
            printf("The name of the patient is %s\n", list->name);
            list = list->next;
        }
    }
    
}

The rest of the work is on progress.

Yes, you should set it to NULL.

When you use malloc, there is no guarantee what value is in the memory you just allocated. It could be anything. That means that if you add new_patient to the end of the list, and you didn't set the next pointer to NULL, it would probably point to some arbitrary piece of memory and you wouldn't have an end to your list. Furthermore, the arbitrary address in new_patient->next is likely memory you don't have access to, so trying to access it would result in undefined behavior. See here for more info on undefined behavior.

I got two more functions but i am a little lost where i should use free function.

Function 1 - Take care of the first client on the list.

Code:

node * see_patient(node * list) {


    node * current;


    current = list;


    // Going to remove the frist patient of the list.
    if (list == NULL) {
        puts("The list is empty");
    } else {
        printf("I am going to see the patient with the name ->: %s\n", list->name);
        list = current->next;
        free(current);
    }


    return list;
}

Function 2 - The patient is tired of waiting and going to give up. Once again i am little bit lost where i should use free function.

Code:

node * tired_of_waiting(node * list) {


    node * current = list;
    node * before = NULL;
    char name[SIZE];


    if (list == NULL) {
        printf("The list is empty.\n");
        return list;
    } else {
        printf("What is the name of the patient that is tried to waitting and going to leave ?\n");
        scanf(" %[^\n]", name);
    }
    getchar();


    // Put the pointers in place.
    while (current != NULL && strcmp(current->name, name) != 0) {
        before = current;
        current = current->next;
    }


    if (current == NULL) {
        puts("That patient is not waitting");
        return list;
    } else {
        if (before == NULL) { // The first on the list
            list = current->next;
        } else {
            before->next = current->next;
        }


        printf("The client with the name %s, was tired of waiting and went away.", current->name);
        free(current); // Is important here ? should i free "before" too ?
        return list;
    }


}