Question

Hi I ran in to this situation. I am using malloc to give me an array of 10 pointers. When I see the test pointers in gdb, one of them(the third )points to 0x0. Sometimes the code segfaults when using apple[2]->string = “hello”. Why does malloc do this? Thanks in advance for the help.

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


int
main(void)
 {
  typedef struct test
    {
      char *string;
      int data;
    } test;

   test *apple[10];  // Declare an array of 10 test pointers. This line results in one of  the  pointers having a null value.
   apple[0] = malloc(sizeof(test));
   apple[0]->string = "hello";

   printf("The string is %s\n",apple[0]->string);
   printf("Size of apple[0]->data is %d\n",sizeof(apple[0]->data));
   printf("Size of tester is %d\n",sizeof(test));
   free(apple[0]);
   return 0;

 }

I wanted to see how the array of pointers would work. I was not intending on using all the 10 pointers. So do I need to malloc only what I need? Is it a coincidence, that the third pointer was 0x0?

Answer 1

Memory has only been allocated for first element in apple so only apple[0] points to a valid struct test.

To allocate memory for all elements of apple:

for (int i = 0; i < sizeof(apple) / sizeof(test*); i++)
{
    apple[i] = malloc(sizeof(test));
}

Similar loop required to free().

test.string is a char*, so pointing to a string literal as you have done is fine (though type should be const char*). If you wish to copy a string to test.string then you must malloc() space to copy into and free() it later.