2

Im getting a segmentation fault when trying to display an element of type (int)

template <class T>
void Lista<T>::imprimir()
{
    NodoL *ptr = new NodoL;
    ptr->sig = pri->sig;
    cout << *ptr->sig->elem; //THIS DISPLAYS CORRECTLY
    cout << *ptr->sig->sig->elem; //SEGMENTATION FAULT
}
HoNgOuRu
  • 717
  • 4
  • 13
  • 26
  • Post the relevant code please. What is `sig`? The answer is "you did it wrong", but it's impossible to tell because you aren't showing the code with the error. – Kerrek SB Sep 07 '11 at 17:15
  • I'm guessing `sig` is some abbreviation of "siguiente" (or somesuch), Spanish for "next", and this is code for a linked list. But please clarify. – R. Martinho Fernandes Sep 07 '11 at 17:17
  • You don't give enough information about what the structure of class NodoL is nor what its constructor does. What is NodoL->sig supposed to be initialised to? This is most likely caused by improper initialisation of the class members in the constructor. – Jon Trauntvein Sep 07 '11 at 17:19
  • Why does your list-printing function create a *new* list node? I'm sure your instructor will tell you that *printing* a list should not modify the list at all. It should not create anything new. And it should probably have a loop so you can print lists of arbitrary size. – Rob Kennedy Sep 07 '11 at 17:24

3 Answers3

7

Are you sure sig is not NULL ?

template <class T>
void Lista<T>::imprimir()
{
    NodoL *ptr = new NodoL;
    ptr->sig = pri->sig;
    cout << *ptr->sig->elem; //THIS DISPLAYS CORRECTLY
    if(ptr->sig == NULL || ptr->sig->sig == NULL)
       return;

    cout << *ptr->sig->sig->elem; //SEGMENTATION FAULT
}
Vivek Goel
  • 22,942
  • 29
  • 114
  • 186
  • with your simple check it returned....that means Im doing something wrong when hooking Nodes previous to this... thanks for the quick response – HoNgOuRu Sep 07 '11 at 17:22
  • well, I found the error when adding the second node to the list I used to have: ult->sig = nuevoNodo; but I changed it to ult->sig->sig = nuevoNodo; ult->sig = nuevoNodo; and now it works as expected. – HoNgOuRu Sep 07 '11 at 17:47
  • nice . if you are using linux you can enable core dump. so you can use gdb to find out where your program had crashed – Vivek Goel Sep 08 '11 at 04:44
1

It appears you have a linked list, where sig points to the next element of the list. Your code allocates a new node and makes it point at the tail of the existing node in pri. If your list was only two elements long to start with, then this code naturally crashes when you attempt to print the third element because there's no such thing. The first element is *ptr->elem, and the second is *pri->sig->elem.

Rob Kennedy
  • 161,384
  • 21
  • 275
  • 467
0

Make sure that elem is a pointer that can be dereferenced, and that it's not either pointing to some invalid location in memory, or that it's not NULL. It seems that you have some type of linked list, and you are attempting to access a list-node that is two nodes away from the current node pointed to by ptr. Either that node may not exist, and therefore sig is an invalid pointer, or the node member elem is an invalid pointer. Either way, you should definitely check pointers before you attempt to dereference so many steps. In fact, this may best be done with something like a for loop such as:

NodoL* temp = ptr;

for (int i=0; i < NUMBER; i++)
{
    if (temp->sig == NULL)
        break;

    temp = temp->sig;
}

cout << *temp->elem << endl;

That way you will either pass through a certain NUMBER of pre-specified nodes in the list from where you're currently at, or you will terminate the for-loop early because you've reached the end of the list.

Jason
  • 31,834
  • 7
  • 59
  • 78