How to malloc (char **)

[Stack Overflow]

Hi all,

I am having some trouble with using malloc() and a (char **). Here is a look at my 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,
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[Dave_Sinkula]

Something like this?

#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.

[Stack Overflow]

Thanks,

That's exactly what I was looking for

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