segmentation fault 11, while pointer variables - c

Question

I am trying to take input from console and add it to hash table.
But I'm getting Segmentation fault 11.
So, I debugged the program using gdb-apple.
It is showing that I'm trying access memory I cannot, using the pointer variable.
I think it is something obvious, but I'm missing it
This is what the gdb is displaying
Program received signal EXC_BAD_ACCESS, Could not access memory.
Reason: KERN_INVALID_ADDRESS at address: 0x0000000000000008
0x0000000100000986 in CreateHashTable (size=200) at hashing.c:29
29 h->Table[i]->next = NULL;
Here is the code
Header File:
#define LOAD_FACTOR 20
#define INITIAL_SIZE 200
struct HashTable *CreateHashTable(int size);
int HashSearch(struct HashTable *h,int data);
int HashInsert(struct HashTable *h,int data);
int HashDelete(struct HashTable *h, int data);
void Rehash(struct HashTable *h);
int Hash(int data, int size);
struct ListNode
{
int key;
int data;
struct ListNode *next;
};
struct HashTableNode
{
int bcount;
struct ListNode *next;
};
struct HashTable
{
int tsize;
int count;
struct HashTableNode **Table;
};
Implementation file:
#include "hashing.h"
#include<stdio.h>
#include<stdlib.h>
struct HashTable *CreateHashTable(int size)
{
struct HashTable *h;
h = (struct HashTable *) malloc ( sizeof(struct HashTable) );
if(h == NULL)
{
printf("Memory Error");
return NULL;
}
h->tsize = (int) size/LOAD_FACTOR;
printf("h->tsize = %d",h->tsize);
h->count = 0;
h->Table = malloc ( ( sizeof(struct HashTableNode **) ) * (h->tsize) );
if( h->Table == NULL )
{
printf("Memory Error");
return NULL;
}
int i;
for( i=0 ; i < (h->tsize) ; i++)
{
h->Table[i]->next = NULL;
h->Table[i]->bcount = 0;
}
return h;
}
I would paste the rest of file, or Driver file, but I don't see it relevant.
Please tell me why I'm getting the segmentation fault 11

Answer 1

You allocated memory for array of pointers but you didn't allocate memory for members of this array.
for( i=0 ; i < (h->tsize) ; i++)
{
h->Table[i] = malloc(...); //put correct arguments here and check allocation
h->Table[i]->next = NULL;
h->Table[i]->bcount = 0;
}

Answer 2

Your problem is here:
struct HashTableNode **Table;
You want an array of nodes (not a 2d array), change to:
struct HashTableNode *Table;
also change
h->Table = malloc ( ( sizeof(struct HashTableNode **) ) * (h->tsize) );
to
h->Table = malloc(sizeof(struct HashTableNode) * h->tsize);
I think I want an array of pointers to nodes, don't I?
As pointed out by #WhozCraig, there is no reason for the additional level of indirection.
Example A (Pointer):
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int *a; /* pointer */
int i, n = 10;
a = malloc(n * sizeof(int)); /* space for 10 ints */
for (i = 0; i < n; i++) {
a[i] = i;
}
for (i = 0; i < n; i++) {
printf("%d\n", a[i]);
}
free(a);
return 0;
}
Example B (Pointer to pointer):
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int **a; /* pointer to pointer*/
int i, n = 10;
a = malloc(n * sizeof(int *)); /* space for 10 pointer to ints */
for (i = 0; i < n; i++) {
a[i] = malloc(sizeof(int)); /* space for 1 int */
*a[i] = i;
}
for (i = 0; i < n; i++) {
printf("%d\n", *a[i]);
free(a[i]);
}
free(a);
return 0;
}
As you can see both do the same thing, but the first one requires less memory and the code is cleaner.
One way to make it easy to remember is:
int * can hold an array
int ** can hold a table (NROWS * NCOLS)
int *** can hold an array of tables