How to malloc (char **)

Hi all,

I am having some trouble with using malloc() and a (char **). Here is a look at my code:

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Words {
        char **wordList;
};

int main() {
        int l;
        int listCount = 4;
        char words[5][25] = {"Season", "Fan", "Car", "Fruit"};
        Words *filter;

        filter = (Words*) malloc ( sizeof(Words) );

        // Write bad words to char array
        for (l = 0; l < listCount; l++) {
                filter->wordList[l] = (char *) malloc (strlen(words[l]) + 1);

                strcpy( filter->wordList[l], words[l] );
        }

        printf("%s\n", filter->wordList[0]);

        for (l = 0; l < listCount; l++) {
                free((char *)filter->wordList[l]);
        }
        free((Words*)filter);

        return 0;
}

---

I'm trying to stay away from using an array of any sort like a char *[], I was just hoping there was a way to allocate memory to a (char **). This is also C, so yes, I would have used new and delete in a split second if it were C++.


Thank you for your time,
- Stack Overflow

Something like this?

Code:

---

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct Words
{
  char **wordList;
};

int main(void)
{
  int i, listCount = 4;
  char words[5][25] = {"Season", "Fan", "Car", "Fruit"};
  struct Words filter;

  filter.wordList = malloc(listCount * sizeof *filter.wordList);
  if ( filter.wordList )
  {
      for ( i = 0; i < listCount; i++ )
      {
        filter.wordList[i] = malloc(strlen(words[i]) + 1);
        if ( filter.wordList[i] )
        {
            strcpy( filter.wordList[i], words[i] );
            printf("%s\n", filter.wordList[i]);
        }
      }
      for ( i = 0; i < listCount; i++ )
      {
        free(filter.wordList[i]);
      }
      free(filter.wordList);
  }
  return 0;
}

/* my output
Season
Fan
Car
Fruit
*/

---

>This is also C

Not really.

Thanks,

That's exactly what I was looking for :)

- Stack Overflow