How to set all values to NULL struct array within a struct array in C

Question

Basically I have two structs and I want to make an array of A structs, and within each A struct I want an array of 50 B structs. So I assume that we will use double pointers.

struct A{
    char* a_word;
    struct B** b_list;
};

struct B{
   char* b_word;
   int b_value;
};

When call initialize function I initialize the structs like this. My goal is to set all the values to NULL when I allocate memory.

struct Word** initialize()
{
    int k;
    int i; 
    struct A** A_list = calloc(BUFFSIZE, sizeof(struct A*));
    for(k =0; k < BUFFSIZE; k++)
    {
     A_list[k] =  calloc (1, sizeof(struct A)); 
     A_list[k]-> b_list =  calloc(50, sizeof(struct B*));

        for(i = 0; i < 50; i++)
        {
           A_list[k]->b_list[i] =  calloc(1, sizeof(struct B)); 
        }

      }
return hashTable;
}

after initializing all these values I am able to do. . .

if(A[43]->a_word == NULL) //43 unallocated value
  {
    printf("This is null\n");
    //program prints this statement - good
    //program knows that this value is NULL
  }

But I also want . .

if(A_list[44]->b_list[0] == NULL)
{
   printf("This is also null");
   //This should be printed but nothing happens
}

For some reason not matter if I set the above if statement to == NULL or != NULL the program outputs absolutely nothing from that if statement. What is going on in memory and how can I allocate everything correctly, and so the value is set to NULL as a default and so I can input a value?

EDIT: Also whenever try to do A_list[value1]->b_list[value2] = strdup("string"); I get a segmentation error, this most likely stems from the same problem.

What do you think this is doing: A_list[k]->b_list[i] = calloc(1, sizeof(struct B)); ? Hint: it is not assigning NULL to that pointer. So testing for one later isn't going to do you much good. Now, what problem is this model supposed to be solving, a two-level hash table? — WhozCraig
– WhozCraig, Commented Apr 26, 2014 at 7:00
Yes I am making a two level hash table. In that calloc statement I am assuming that I am allocating space for a B struct. — user3312266
– user3312266, Commented Apr 26, 2014 at 7:02
...which means checking A_list[44]->b_list[0] == NULL will evaluate as false, since you're initialization loop clearly populates it with an allocation address (assuming calloc doesn't fail, which should also be checked). Testing for non-null and fixing the output flush as alk's answer shows should get you output. — WhozCraig
– WhozCraig, Commented Apr 26, 2014 at 7:06
Your posted code won't compile. You don't have the declaration of hashTable. You should post the actual code that you are having problems with. — R Sahu
– R Sahu, Commented Apr 26, 2014 at 7:09

Community · Accepted Answer · 2017-05-23 11:49:25Z

4

As mentioned already by WhozCraig in a comment to the question, this code

    for(i = 0; i < 50; i++)
    {
       A_list[k]->b_list[i] =  calloc(1, sizeof(struct B));

initialises the first 50 elements of b_list to point to valid memory, that is to be non 0, assuming calloc() never fails. Being that optimistic you better test for those elements being != NULL.

... if I set the above if statement to == NULL or != NULL the program outputs absolutely nothing

The code does not seem to flush stdout here:

if(A_list[44]->b_list[0] == NULL)  
{
   printf("This is also null");

Change this by adding a final \n:

if(A_list[44]->b_list[0] != NULL)  
{
   printf("This isn't null\n");

As stdout is line buffered by default, all content will be flushed if a new-line is detected.

edited May 23, 2017 at 11:49

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answered Apr 26, 2014 at 7:00

alk

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