Thread: storing dynamically allocated pointers

yes, the names have to be reachable to other functions called in main, i tryed to do something like

tab[i] = (char*) malloc (sizeof (char) * strlen(name));
printf("%s",(*tab)[i]);

just to check if it was stored succesfully, but hen a processor fault came up.

tab[i] = (char*) malloc (sizeof (char) * strlen(name));
printf("%s",(*tab)[i]);

The following line will allocate enough memory for your string. Note that a string always uses one char more than it's string length for the terminating null character.

tab[i] = malloc( sizeof (char) * (strlen(name)+1) );

Keep in mind that malloc does not in any way prepare the data. You have not copied anything to tab[i] yet, so the next line, while syntactically correct, might crash your system or print garbage if you're lucky.

printf("%s", tab[i]);

You need to actually copy the contents of the string before printing:

strcpy( tab[i], name );

ollright it works, but how come you dont have to use a dereference operator to access the content the pointers in the array refer to?

Did you have a question?

If you don't write to the memory allocated, how is the memory you allocated supposed to contain the data you would like it to contain? [edit]And leaking the memory you allocated by reassigning the pointer to a local array is not going to help. Understand that arrays cannot be assigned.

Originally Posted by kris.c

.. If that is the case, then fputs should print junk inside the function "input " itself .. Its printing whatever I enter correctly .. That is some proof that , it is being stored in the location pointed to by names[i]..

No, it just means that you reassigned the pointer that had previously been given whatever malloc returned to instead point to the local array. When this array was in scope and contained a string, fputs could of course correctly print the string being pointed to.

Originally Posted by kris.c

That means ,the array is containing the pointers to the strings, right??
so, if the issue is only with scope of the array , will making the array global help ?

Let's not go there. How about we learn how to do this right?

Code:

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

void input(char *array[], size_t size)
{
   size_t i;
   for ( i = 0; i < size; i++ )
   {
      char name[21];
      printf("enter text #%d of %d: ", (int)i + 1, (int)size);
      fflush(stdout);
      if ( fgets(name, sizeof name, stdin) != NULL )
      {
         char *newline = strchr(name, '\n');
         if ( newline != NULL )
         {
            *newline = '\0';
         }
         array[i] = malloc(strlen(name) + 1);
         if ( array[i] != NULL )
         {
            strcpy(array[i], name);
         }
         printf("array[%d] = \"%s\"\n", (int)i, array[i]);
      }
   }
}

int main (void)
{
   char *names[3];
   size_t i;
   input(names, sizeof names / sizeof *names);
   for ( i = 0; i < sizeof names / sizeof *names; i++ )
   {
      printf("names[%d] = \"%s\"\n", (int)i, names[i]);
   }
   return 0;
}

/* my output
enter text #1 of 3: one
array[0] = "one"
enter text #2 of 3: two
array[1] = "two"
enter text #3 of 3: three
array[2] = "three"
names[0] = "one"
names[1] = "two"
names[2] = "three"
*/

Originally Posted by kris.c

I appreciate that.. My prog is working perfectly with strcpy() .. I wanted to know why things fail with the other method.. PArticularly because , the contents of the array are valid inside the function ..

The "contents" of the array are never valid anywhere because you haven't put any contents into the array -- you've merely pointed elsewhere.

[edit]Why not post what you think ought to work, and perhaps it will be easier to explain why not.