How to dynamically allocate memory for every element in 2D array in C

Question

how to dynamically allocate memory for every element in 2D array? For example, I have an address which consist of street name(1.column) and street number(2.column) e.g. St.Paul 87/45. And I need to dynamically allocate memory for them.

int main()
{
   int c = 1, r = 1; //c-column, r-row
   char st_name[] = "St.Paul";
   char st_number[] = "87/45";
   char *arr = (char *)malloc(c*r*sizeof(char));
   c = 0; 
   r = 0;
   *arr[r][c++] = (char *)malloc((strlen(st_name)) * sizeof(char));
   *arr[r][c] = (char *)malloc((strlen(st_number)) * sizeof(char));
   return 0;
}

Of course it´s not working.

Thanks.

[Please see this discussion on why not to cast the return value of `malloc()` and family in `C`.](http://stackoverflow.com/q/605845/2173917). — Sourav Ghosh, Mar 17 '17 at 10:00
This is not going to work that way in a straightforward approach, you'll need at least a pointer to an array. — Sourav Ghosh, Mar 17 '17 at 10:02
Souraw Ghosh show him also link why do cast return value of malloc. — unalignedmemoryaccess, Mar 17 '17 at 10:03
Solution here belongs to 3 stars variable, like [this example](http://ideone.com/SmCst9) — LPs, Mar 17 '17 at 10:10
@tilz0R: The post claiming that casting `malloc` makes your code more portable between C and C++ with 75 downvotes? — David Ranieri, Mar 17 '17 at 10:14
http://stackoverflow.com/a/14879184/3716664 Could not say 75 downvotes. @KeineLust — unalignedmemoryaccess, Mar 17 '17 at 10:15
@tilz0R, you need more rep., 5000 or 7000, I don't remember ... — David Ranieri, Mar 17 '17 at 10:29

LPs · Accepted Answer · 2017-03-17T10:29:05.940

What you need is a matrix, but each matrix item is a c-string so an array.

You can do it using 3 starts pointer, like:

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

int main(void)
{
   int c = 1, r = 1; //c-column, r-row
   char st_name[] = "St.Paul";
   char st_number[] = "87/45";
   char ***arr = malloc(r*sizeof(char *));
   c = 0;
   r = 0;
   if (arr != NULL)
   {
       arr[r] = malloc(c*sizeof(char *));

       if (arr[r] != NULL)
       {
           arr[r][c] = malloc(strlen(st_name)+1);
           if (arr[r][c] != NULL)
           {
               c++
               arr[r][c]   = malloc(strlen(st_number)+1);
               if (arr[r][c] != NULL)
               {        
                   sprintf( arr[0][0], st_name);
                   sprintf( arr[0][1], st_number);

                   printf ("arr[0][0] = %s\n", arr[0][0]);
                   printf ("arr[0][1] = %s\n", arr[0][1]);
               }
           }
       }
   }

   return 0;
}

Or you can do that using a struct that defines each item of your array, like

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

struct item
{
    char *name;
    char *number;
};

int main(void)
{
   char st_name[] = "St.Paul";
   char st_number[] = "87/45";
   int r = 1;

   struct item  *arr = malloc(r*sizeof(struct item));

   if (arr != NULL)
   {
       arr[0].name = malloc(strlen(st_name)+1);
       arr[0].number  = malloc(strlen(st_number)+1);

       if ((arr[0].name != NULL) && (arr[0].number != NULL))
       {
           sprintf( arr[0].name, st_name);
           sprintf( arr[0].number, st_number);

           printf ("arr[0][0] = %s\n", arr[0].name);
           printf ("arr[0][1] = %s\n", arr[0].number);

           free(arr[0].name);
           free(arr[0].number);
       }

       free(arr);
   }

   return 0;
}

Thank you so much. It help a lot! – KarlJohanson Mar 17 '17 at 10:29 — KarlJohanson, Mar 17 '17 at 10:29

How to dynamically allocate memory for every element in 2D array in C

1 Answers1