C list of a list - c

Question

im having a problem, im trying to make a list of a list where inicioC refers to the first node of clients, and every node of clients will have a list of rentals, referred as inicioA. The thing is, i dont know how to save the first pointer of the rentals, like, the first node of clients will only be saved one time, but wont the first node of every rental be different?
this are the structs:
typedef struct _aluguer
{
int id, estado, diaI, mesI, anoI, diaE, mesE, anoE;
struct _aluguer *prox;
}aluguer, *pAluguer;
typedef struct _registo
{
int nif,nalu;
char nome[100];
struct _registo *prox;
pAluguer *inicioA;
}registo, *pRegisto;
and this is the code I use to extract the info from a file into the list of lists
pRegisto iniC(pRegisto inicioC, pAluguer inicioA)
{
FILE *c;
int j, nif, nalu, l=0;
char nome[100];
c = fopen("clientes.txt", "r"); // abrir ficheiro
if(c == NULL)
{
printf("Erro ao abrir ficheiro %s", "clientes.txt");
exit(0);
}
while(fscanf(c, "%d %d %s", &nif, &nalu, nome) == 3) //format of info
{
pRegisto novoC = malloc(sizeof(registo));
if(novoC == NULL)
{
printf("erro alocacao memoria\n");
return inicioC;
}
novoC -> prox = NULL;
pAluguer inicioA = NULL;
pRegisto aux = inicioC;
pRegisto p = NULL;
novoC->nif=nif;
novoC->nalu=nalu;
strcpy(novoC->nome, nome);
while(aux != NULL)
{
p = aux;
aux = aux->prox;
}
if( aux == inicioC)
{
inicioC=novoC;
}
else
{
p->prox = novoC;
}
for(j=0; j<novoC->nalu; j++) // repeat is equal to the number of rentals
{
l++;
pAluguer novoA = malloc(sizeof(aluguer));
if(novoA == NULL)
{
printf("erro alocacao memoria\n");
return inicioC;
}
novoA -> prox = NULL;
pAluguer aux = inicioA;
pAluguer p = NULL;
fscanf(c, "%d %d", &(novoA->id), &(novoA->estado));
if(novoA->estado == 0)
{
fscanf(c, " %d %d %d", &(novoA->diaI), &(novoA->mesI), &(novoA->anoI));
}
else
{
fscanf(c, " %d %d %d %d %d %d", &(novoA->diaI), &(novoA->mesI), &(novoA->anoI), &(novoA->diaE), &(novoA->mesE), &(novoA->anoE));
}
while(aux != NULL)
{
p = aux;
aux = aux->prox;
}
if( aux == inicioA)
{
inicioA=novoA;
}
else
{
p->prox = novoA;
}
}
}
fclose(c);
return inicioC;
}

Answer 1

Arraylist (Array of pointers in C) is good choice for implementing list of list but as you are forced to do in this way, I am showing a Simple method how you can develop list of list in C.
It is very simple,suppose you have a list of 3 numbers: 1,2,3 i.e list_1,list of 2 numbers: 5,6 i.e list_2.Now these list_1 and list_2 can be linked in a similar way as the numbers in list linked above.Look at this scenario
list_1:1->2->3->Null
list_2:5->6->Null
list_of_list : list_1->list_2->Null i.e. (1->2->3->Null) -> (5->6->Null) -> Null
Here is a Sample program to insert list of ints in Last:
#include <stdio.h>
#include <stdlib.h>
// for list of Intergers
struct node{
int Data;
struct node *next;
};
// this is for list of nodes i.e list of lists
struct list{
struct node *start;
struct list *listnext;
};
// insert integers in list
void insertNode(struct node **head ,int data){
struct node *temp,*current;
temp=malloc(sizeof(struct node));
temp->Data=data;
temp->next=NULL;
if((*head)==NULL){
(*head)=temp;
}
else{
current=(*head);
while(current->next!=NULL){
current=current->next;
}
current->next=temp;
}
}
// insert lists of integers in a list
void insertList(struct list **Listhead,struct node *head){
struct list *temp,*current;
temp=malloc(sizeof(struct list));
temp->start=head;
temp->listnext=NULL;
if((*Listhead)==NULL){
(*Listhead)=temp;
}
else{
current=(*Listhead);
while(current->listnext!=NULL){
current=current->listnext;
}
current->listnext=temp;
}
}
// Show all the list with their data
void show(struct list *Listhead){
int i=1;
struct list *current;
struct node *currentlist;
current=Listhead;
while(current!=NULL){
currentlist=current->start;
printf("List %d: ",i);
while(currentlist!=NULL){
printf("%d ",currentlist->Data);
currentlist=currentlist->next;
}
i++;
printf("\n");
current=current->listnext;
}
}
int main(){
struct node *head1=NULL,*head2=NULL,*head3=NULL; // 3 lists of integers
struct list *Listhead=NULL; // Listhead will be the head of lists of list
insertNode(&head1,20); // inserting in first list
insertNode(&head1,13);
insertNode(&head1,22);
insertNode(&head1,18);
insertNode(&head2,42); // inserting in second list
insertNode(&head2,15);
insertNode(&head3,12); // inserting in third list
insertNode(&head3,14);
insertNode(&head3,28);
insertList(&Listhead,head1); // inserting lists in list
insertList(&Listhead,head2);
insertList(&Listhead,head3);
show(Listhead);
}
Output for this Program:
List 1: 20 13 22 18
List 2: 42 15
List 3: 12 14 28
I hope you got the point.