C List Segmentation Fault - c

Question

Hello guys i got a problem while running this code:
trying to run the printf in the comment i got a segfault, also without that i dont see my listed printed ad the function Stampa should do.
Probably i am missing something with pointers
#include <stdio.h>
#include <stdlib.h>
struct nodo { // double linked list
int info;
struct nodo *prec;
struct nodo *succ;
};
typedef struct nodo nodo;
struct nodo *Crealista(void);
void Stampa (struct nodo *nodo);
int main(int argc, const char * argv[]) {
struct nodo *p;
p = Crealista();
// printf("%d",p->succ->info);
Stampa(p);
return 0;
}
// this funct should print the whole list
void Stampa (struct nodo *p) {
while (p->succ != NULL ) {
printf("Value : %d \n", p->info);
p = p->succ;
}
}
// this funct should create list with n members and return a pointer to the first element
struct nodo *Crealista(void) {
struct nodo *p, *primo, *back;
int i, n;
p = NULL;
primo = NULL;
back = NULL;
printf("Numero di elementi: ");
scanf("%d", &n);
printf("inserisci %d numeri interi positivi: ", n);
for (i=0; i<n; i++) {
p = malloc(sizeof(struct nodo));
scanf("%d", &p->info);
p->prec = back;
p->succ = NULL;
back = p;
}
primo = p;
while (primo->prec != NULL) { primo = primo->prec;}
return(primo);
}

Answer 1

Stampa()
Let's look at what we want to do when we print the entire list:
If the current element is valid, then we want to print it.
Then, we want to iterate to the next element, and continue this loop.
That's not what your code does. Your code looks to see if the current element has a successor, and if it does, we print the current element's value.
That function should actually be:
void Stampa (struct nodo *p) {
while (p != NULL ) {
printf("Value: %d\n", p->info);
p = p->succ;
}
}
Crealista()
If you tell Crealista() to create a list of 0 elements, your final while loop will exhibit undefined behavior.
If you tell Crealista() to create a list of less than 2 elements, your commented printf() in main() will cause undefined behavior (if it was uncommented).
Doubly Linked Lists
You never update the value of nodo->succ. You only update the value of nodo->prev. Here's one example of how to do that:
if (p->prec)
p->prec->succ = p;
Putting all of this together
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
struct nodo {
int info;
struct nodo *prec;
struct nodo *succ;
};
typedef struct nodo nodo;
struct nodo *Crealista(void);
void Stampa (struct nodo *nodo);
int main(int argc, const char * argv[]) {
struct nodo *p;
p = Crealista();
Stampa(p);
}
// Print the list
void Stampa (struct nodo *p) {
while (p != NULL ) {
printf("Value: %d \n", p->info);
p = p->succ;
}
}
// Create a list of n elements
struct nodo *Crealista(void) {
int i, n;
struct nodo *p = NULL;
struct nodo *primo = NULL;
struct nodo *back = NULL;
printf("Numero di elementi: ");
scanf("%d", &n);
assert(n != 0);
printf("inserisci %d numeri interi positivi: ", n);
for (i=0; i<n; i++) {
p = malloc(sizeof(struct nodo));
scanf("%d", &p->info);
p->prec = back;
p->succ = NULL;
if (p->prec)
p->prec->succ = p;
back = p;
}
primo = p;
while (primo != NULL && primo->prec != NULL)
primo = primo->prec;
return primo;
}
Which when run...
Numero di elementi: 5
inserisci 5 numeri interi positivi: 1 2 3 4 5
Value: 1
Value: 2
Value: 3
Value: 4
Value: 5

Answer 2

The printf() call you have commented out will always exhibit undefined behavior when p->succ is NULL, as it will be when p points to the last (or only) element of the list. In that case, although the behavior is formally "undefined", it is quite likely to manifest as a segmentation fault.

Answer 3

In stampa add check
while(p!=NULL)
{
printf("Value %d",p->info);
p=p->succ;
}
You are probably accessing it when it is not allocated. Check it this way there is no problem if it reaches a null value also. And initial null value will not cause problem also(if used).
I have modified your code
// this funct should create list with n members and return a pointer to the first element
struct nodo *Crealista(void) {
struct nodo *p, *primo, *back;
int i, n;
p = NULL;
primo = NULL;
back = NULL;
printf("Numero di elementi: ");
scanf("%d", &n);
printf("inserisci %d numeri interi positivi: ", n);
p = (struct nodo *)malloc(sizeof(struct nodo));
scanf("%d", &p->info);
p->prec = back;
p->succ = NULL;
back = p;
for (i=1; i<n; i++) {
p = (struct nodo *)malloc(sizeof(struct nodo));
scanf("%d", &p->info);
p->prec = back;
p->prec->succ=p;
p->succ = NULL;
back = p;
}
primo = p;
while (primo->prec != NULL) { primo = primo->prec;}
return(primo);
}
You are not setting the succ pointer of previous node correctly. That is what causing you the problem. Check now.