Linked List error - C

Question

I'm just messing around with linked lists but it seems I have many things wrong in my mind. I've tried this with functions and I couldn't get it working. I thought I'd get it right by making it simplier but it still doesn't work. It usually prints just the first element right and then random huge numbers (memory addresses most likely)

I just want to create a list and print its content.

#include <stdio.h>
#include <stdlib.h>

struct el{
    int value;
    struct el *next;
};

typedef struct el Elem;



int main()
{
    int nr, i;
    struct el *Head, *Conductor;
    Head = malloc(sizeof(Elem));
    Conductor = Head;
    printf("How many elements do you want to add? ");
    scanf("%d", &nr);
    for(i = 0; i < nr; i++)
    {
        printf("Enter value for element %d:  ", i);
        scanf("%d", &(Conductor->value));
        Conductor = Conductor->next;
        Conductor = malloc(sizeof(Elem));


    }
    free(Conductor->next);
    Conductor->next = NULL;

    Conductor = Head;
    printf("\n");

    for(i = 0; i < nr; i++)
    {
        printf("%d ->  ", Conductor->value);
        Conductor = Conductor->next;
    }

   return 0;
}

First allocate memory, then assign the value.

0xc0de
– 0xc0de

2013-05-31 12:06:05 +00:00
Commented May 31, 2013 at 12:06 — 0xc0de
– 0xc0de, Commented May 31, 2013 at 12:06

sirius · Accepted Answer · 2013-05-31 12:17:06Z

6

You should allocate memory for Conductor->next before assigning Conductor->next to Conductor.

for(i = 0; i < nr; i++)
{
    printf("Enter value for element %d:  ", i);
    scanf("%d", &(Conductor->value));
    Conductor->next = malloc(sizeof(Elem));
    Conductor = Conductor->next;
}

It's important that you make the allocation before the assignment because otherwise the "old" Conductor->next will not point to the new Conductor.

edited May 31, 2013 at 12:17

answered May 31, 2013 at 12:05

sirius

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Comments

Some programmer dude · Accepted Answer · 2013-05-31 12:08:40Z

5

You do not allocate memory for the next pointer.

You need to allocate before assigning:

scanf("%d", &(Conductor->value));
Conductor->next = malloc(sizeof(Elem));
Conductor = Conductor->next;

edited May 31, 2013 at 12:08

answered May 31, 2013 at 12:02

Some programmer dude

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2 Comments

Arlind Over a year ago

I did a mistake while copying the code. Check it now. Is this allocation right?

Some programmer dude Over a year ago

@Arlind No, you're doing it in the wrong order, please see my updated answer.

Deepu · Accepted Answer · 2013-05-31 12:05:55Z

2

The memory should be allocated to Conductor->next,

Change the loop as follows,

for(i = 0; i < nr; i++)
{
    printf("Enter value for element %d:  ", i);
    scanf("%d", &(Conductor->value));
    Conductor->next = malloc(sizeof(Elem));
    Conductor = Conductor->next;

}

answered May 31, 2013 at 12:05

Deepu

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Comments

Nobilis · Accepted Answer · 2013-05-31 12:16:51Z

Right here: Conductor = Conductor->next;

Conductor is assigned a NULL value, so when you try to dereference it to store input value in the next iteration of the loop you get a segmentation fault.

As others have pointed out you should first allocate memory for Conductor->next using malloc.

Can I finally recommend that you have a look at a debugger such as gdb as it makes identifying such problems a breeze?

This is how I found out that there was a problem with Conductor->next using gdb:

I stepped through the program until I reached the offending instruction:

25    Conductor = Conductor->next;

I tried to print what's in next:

(gdb) print Conductor->next

And here's what I got:

Cannot access memory at address 0x4

Whereas this is what I have in Conductor

(gdb) print Conductor
$7 = (struct el *) 0x804b008

Very easy to see we have a problem as in the next loop it will be assigned an invalid value;

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