storing dynamically allocated pointers

hi still struggling with pointers here,i want to put in names and then dynamically allocate memory to store the names. But how do i store he pointer given by malloc into the array of pointers.

Code:

---

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

void input (char *array[]){

        char name[21];                                                //first storing name into name[], 20 char max

        for(i=0; i<10; i++){
                fgets(name,20,stdin);
                ... = (...) malloc (sizeof (char) * strlen(name));
        }
return
}

int main (void){

        char *names[10];                //room for 10 pointers
        int count;

return 0;
}

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Firstly, Malloc needs no typecasting...
But, is this what u r trying to do here? U first store the input in "name " , then u r allocating space that is as big as " name " and u are storing the pointer that malloc returns in an array of pointers . back , in main, u wish to print each string

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.

malloc returns a void pointer, no need to type cast .. read FAQ
U r allocating the space as large as the string "name", thats fine , u make an element of the array of that array of char pointers u have declared , point to the location , ok, but that location is still not containing anything and u r trying to print it ..
doing

Code:

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  tab[i] = name ;
  fputs( tab[i] , stdout );

---

in ur function might work ,i guess.. just try..

Quote:

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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?

U dont need a de-reference operator because the strcpy takes in arguments that are pointers to chars.. "tab[i]" and "name" are pointers to chars.. The same is the case with fputs..

Now that the problem has been solved, I still have one doubt ..

suppose , in the function, I do,

Code:

---


  tab[i] = name ;
  fputs( tab[i] , stdout );

---

after allocating space for tab[i].. It prints the name alright,
But back in main ,

Code:

---

        for(count=0;count < 3; count++)
          fputs(names[count],stdout);

---

It prints junk in the end..

It works fine with strcpy() as sugggested by nv..

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.

.. 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].. It prints junk only in main..yeah, ok i will check the addresses of each of the element in names[i] inside my function.. And i dont think i have done any mistakes in passing the array to hte function..

Quote:

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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]..

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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.

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 ?

Quote:

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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 ?

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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"
*/

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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 problem required the array now to be seen elsewhere too..

Quote:

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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 ..

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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.

the well seems to getting deeper and deeper...

> arrays cannot be assigned.
so, this means the :

Code:

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  array[i] = name ;

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is itself wrong ?
but, we can assign one pointer to another and use the other pointer for referencing data for normal variables.. The array we have is a set of pointers in itself .. So, I got mislead into the notion that i can assign the "name " pointer into one of the array elements and then do what ever needed..

yeah, ok I shud have done that long time back

Code:

---


  void input ( char *names[]){
        int i;
        char name[21];
       
        for(i=0; i<3; i++)
        {
                printf("enter the name \n");
                fgets(name,20,stdin);

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

                names[i] = name ;
                fputs(names[i],stdout);
                printf("i=%d\n",i);
               
        }
return ;
}

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that is my function ..

In main I just print them out :

Code:

---

          for(count=0;count < 3; count++)
          fputs(names[count],stdout);

---

I just hope ,i havent made any real blunder with my sleepy eyes here..

Code:

---

  void input ( char *names[]){
        int i;
        char name[21];
       
        for(i=0; i<3; i++)
        {
                printf("enter the name \n");
                fgets(name,20,stdin);

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

                names[i] = name ;
                fputs(names[i],stdout);
                printf("i=%d\n",i);
                printf("%u\n",names[i]);
        }
return ;
}

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Do you see the memory leak in bold, and the difference between similar names in different colors? And do you understand what is going on?

yeah, I think I do... better stick to the other method , i guess , instead of messing around with memory like this

If it's still not clear, it seems a bit like wondering what happens to the 42:

Code:

---

#include <stdio.h>

int main (void)
{
  int x;
  x = 42;
  printf("x = %d\n", x);
  x = -1;
  printf("x = %d\n", x);
  return 0;
}

/* my output
x = 42
x = -1
*/

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yeah, thanks dave..