Array of structs, unable to check if entry is NULL

Question

Im toying with creating my own hash table data structure, but I have hit an unexpected problem, which I can't solve and haven't found a satisfying solution for it yet.

See, I have this linked list struct

struct Link
{
   int v;
   struct Link* next;
}

and then inside the hash table structure I want keep track of an array linked lists, like this:

struct Link** entries;

What I run into is, that for this to work, I first have to initialize the array like this:

entries = malloc(sizeof(struct Link*) * N);

for (i = 0; i < N; i++)
   entries[i] = malloc(sizeof(struct Link));

What I want is not having to do the for loop that initializes the structs, because that's not how linked lists work, I want to leave entries[x] blank until it actually gets assigned a value.

If I don't do the for loop this happens:

if (entries[x] != NULL)  /* true, the array is initialized */
  entries[x]->v = value; /* SEGFAULT, there is no struct initialized */

That if statement 'should' return false if I haven't assigned a Link struct to it yet, but it doesn't.

One way of solving this problem is to just initialize all the first links of the linked list with that for loop and then checking for the value, but that's not what I want.

So does anyone know a way to solve this the way I want it to?

Because you don't want to have an 'empty' link on the start of the linked list. — Martndemus
– Martndemus, Commented Sep 1, 2012 at 8:28
you can use calloc instead of using malloc when allocating entries. so your checking extires[x] against NULL works better. — sardok
– sardok, Commented Sep 1, 2012 at 8:38

Rohan · Accepted Answer · 2012-09-01 08:31:46Z

1

You can use calloc rather than malloc while allocating entries like:

entries = calloc(sizeof(struct Link*),  N);

answered Sep 1, 2012 at 8:31

Rohan

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2 Comments

Eitan T Over a year ago

@Rohan As a sidenote, you swapped the calloc parameters, although it doesn't really matter. The first parameter should be N and the second is the size.

Jens Gustedt Over a year ago

Strictly speaking calloc is not the right tool to initialize pointer values. There seem to be architectures out there for which a null pointer is not represented by an all 0 byte pattern. So be a bit careful when using this.

user529758 · Accepted Answer · 2012-09-01 08:27:06Z

1

Well, you can't. Uninitialized pointers are not necessarily NULL.

answered Sep 1, 2012 at 8:27

user529758

Comments

Frank Bollack · Accepted Answer · 2012-09-01 08:38:02Z

0

You will need to initialize the pointer array after you allocate space for it with malloc(). You can use another for-loop like:

for (i = 0; i < N; i++)
{
   entries[i] = NULL;
}

or simply use calloc() to allocate the array. This will set all elements to 0.

entries = calloc(N, sizeof(struct Link*));

Afterwards you need to allocate the elements as needed:

if (entries[x] == NULL)
{
   entries[x] = malloc(sizeof(struct Link));
}
entries[x]->v = value;

answered Sep 1, 2012 at 8:38

Frank Bollack

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