memory allocation in array of structure using double pointers in C

Question

Below is the sample code for populating the array of struct R using double pointers. I am unable to allocate memory to r[0] and also when the function exits, both r and r[0] becomes 0x0.

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

struct R
{
  int x;
  int y;
  char * z;
};

void func(struct R **r)
{
  r = (struct R **) malloc(4 * sizeof(struct R *));
  r[0] = (struct R *) malloc(sizeof(struct R));   // giving r[0] = 0x0
  r[0]->x = 1;
  r[0]->y = 2;
  r[0]->z = (char *) malloc(64 * sizeof(char));
  strcpy(r[0]->z , "HELLO");
}

int main()
{
  struct R *r = NULL;
  func(&r);
  printf("%d", r->x);
  printf("%d", r->y);
  printf("%s", r->z);
  return 0;
}

I am unable to find the reason behind it. Any help would be highly appreciated.

R Sahu · Accepted Answer · 2018-03-30T07:09:47.250

The line

r = (struct R **) malloc(4 * sizeof(struct R *));

changes where r points but only locally in the function. It does not change the value of the pointer in the calling function.

What you need is:

void func(struct R **r)
{
   *r = malloc(sizeof(struct R));

   r[0]->x = 1;
   r[0]->y = 2;
   r[0]->z = malloc(64 * sizeof(char));
   strcpy(r[0]->z , "HELLO");
}

Another option is to change the return value of func to a pointer and make it simpler to use.

struct R * func()
{
   struct R *r = malloc(sizeof(*r));

   r->x = 1;
   r->y = 2;
   r->z = malloc(64 * sizeof(char));
   strcpy(r->z , "HELLO");

   return r;
}

and use it in main as:

struct R *r = func();

PS See Do I cast the result of malloc? to understand why you should not cast the return value of malloc.

An even better recommendation (than the usual "don't cast malloc") would be to use the `sizeof` operator on the actual *variable*, instead of repeating the type again. I.e. `int *x = malloc(sizeof *x)` always works and is a simple rule to remember, while the main problem with casting malloc (implicit `int` function declaration) is no longer an issue as of C11, and not even with earlier compilers with proper warnings enabled. — vgru, Mar 30 '18 at 09:02

David Ranieri · Answer 2 · 2018-03-30T07:07:30.327

You want an array of struct R, then why are you reserving space for an array of pointers to struct R?

Change to:

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

struct R
{
  int x;
  int y;
  char * z;
};

void func(struct R **r)
{
  /* Dereference in order to change the value of r */
  *r = malloc(4 * sizeof(struct R)); /* Don't cast malloc */
  r[0]->x = 1;
  r[0]->y = 2;
  r[0]->z = malloc(64); /* char is always 1 byte */
  strcpy(r[0]->z , "HELLO");
}

int main(void) /* Correct prototype */
{
  struct R *r = NULL;

  func(&r);
  printf("%d", r[0].x);
  printf("%d", r[0].y);
  printf("%s", r[0].z);
  /* Don't forget to free */
  free(r[0].z);
  free(r);

  return 0;
}

score 1 · Answer 3 · answered Mar 30 '18 at 06:43

This line:

void func(struct R **r)
{
  r = (struct R **) malloc(4 * sizeof(struct R *));

No matter what type r is, it's still a local variable. If you assign a local variable, the caller part won't be affected. This code is similar to

void func(int r)
{
  r = 1;
}

memory allocation in array of structure using double pointers in C

3 Answers3