Seg fault when trying to access member of structure array

Question

typedef struct abc {
  unsigned int pref;
  unsigned int port;
  char *aRecordIp;
    int index;
    int count;
}abc_t;

typedef struct xyz {
        abc_t *ab;
        int index;
        int count;
}xyz_t;


int Lookup (char *lookup,xyz_t **handle) {

*handle = (xyz_t *)malloc(sizeof(xyz_t *));
(*handle)->ab = (abc_t *) malloc(5*sizeof(abc_t *));
//
(*handle)->ab[0].pref = 10;  //seg fault here
}


void *myhandle; 

// this is a void pointer and I cannot change that as it is used elsewhere too

char lookup;


Lookup(&lookup, &myhandle);

// I get a warning here which is obviuous but how do I take care of that

Answer 1

(*handle)->ab = (abc_t *) malloc(5*sizeof(abc_t *));

Should probably be:

(*handle)->ab = (abc_t *) malloc(5*sizeof(abc_t));

In C you don't need that cast though, since void* casts implicitly to any kind of pointer, so:

(*handle)->ab = malloc(5*sizeof(abc_t));

is enough.

Also, make sure you check the return value of all malloc calls.

Answer 2

(abc_t *) malloc(5*sizeof(abc_t *)); should be (abc_t *) malloc(5*sizeof(abc_t)); . You are requesting for space of 5 pointers rather than space for 5 strucures.

Answer 3

That error is a good reason to prefer the ptr = malloc (count * sizeof *ptr); notation; the compiler already knows the type and size of an expression, so you won't have to specify that again :

int Lookup (char *lookup,xyz_t **handle) {

  *handle = malloc(sizeof **handle);
  (*handle)->ab = malloc(5 * sizeof *(*handle)->ab );
  (*handle)->ab[0].pref = 10;  //seg fault here
}

Or, using array notation (just to avoid the ugly *(*handle) ) :

int Lookup (char *lookup,xyz_t **handle) {

  *handle = malloc(sizeof **handle);
  (*handle)->ab = malloc(5 * sizeof (*handle)->ab[0] );
  (*handle)->ab[0].pref = 10;  //seg fault here
}

Note: I also removed the casts, which are not needed and potentially dangerous.

Also note that this stylistic correction will fix both the errors.