Learning C
Learning C

Reputation: 689

Why does my whole array have '\0'?

This is causing a lot of issues for my program. Whys is it when I create a new array of structured pointers that they all equal to '\0'? I checked if they are at the end of the array if(table_p -> buckets_array[i] == '\0'){ printf("ask this \n") ; } and it is true for every member of the array. Am I checking it wrong? Shouldn't only the last member have \0?

typedef struct data_{
  char *key;
  void *data;
  struct data_ *next;
}data_el;

typedef struct hash_table_ {
  void **order;
  int *number_next_calls;
  int *number_buckets;
  int *buckets_size;
  int *worst;
  int *total;
  float *average;
  int (*hash_func)(char *);
  int (*comp_func)(void*, void*);
  data_el **buckets_array;
} hash_table, *Phash_table;

/*Create buckets array*/
table_p -> buckets_array = (data_el **)malloc(sizeof(data_el *)*(size+1));
table_p -> buckets_size = (int *)malloc(sizeof(int)*(size+1));

/*Setting order array*/
  table_p -> order = NULL;

/*Setting inital condictions*/
table_p -> worst = (int *)malloc(sizeof(int));
table_p -> total = (int *)malloc(sizeof(int));
table_p -> average = (float *)malloc(sizeof(float));
table_p -> number_buckets = (int *)malloc(sizeof(int));

/*This is where I have isssue*/
for(i = 0; i < size; i++){
    table_p -> buckets_array[i] = NULL;
    table_p -> buckets_array[i] -> buckets_size = 0;

    if(table_p -> buckets_array[i] == '\0'){
      printf("ask this \n");
    }
}

Upvotes: 0

Views: 126

Answers (2)

wildplasser
wildplasser

Reputation: 44250

if(table_p -> buckets_array[i] == '\0'){

I already pointed out in your previous (very similar) post, that table_p->buckets_array[i] is of type "pointer to data_el", and that '\0' is an integer constant. Please read before you post.

Upvotes: 0

Vaughn Cato
Vaughn Cato

Reputation: 64308

In your code, you have:

table_p -> buckets_array[i] = NULL;
table_p -> buckets_array[i] -> buckets_size = 0;

That is like saying:

table_p -> buckets_array[i] = NULL;
NULL -> buckets_size = 0;

which is not good.

Upvotes: 2

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