How to return dynamic array from void function in c?

Question

I want to return a dynamic array by reference from a void function.
I already searching 3 hours for the answer, couldn't find anything helpfull.
Here is my simplified code :

main()

{

    int **a;



    xxx(&a);



    printf("%d\n\n", a[1]);



}



void xxx(int **a)

{

    int i;



    *a = (int*)malloc(5 * 4);



    for (i = 0; i < 5; i++)

        a[i] = i;

    printf("%d\n\n", a[1]);

}

I just want to allocate dynamic array in the "xxx" Function and return it by reference to main, than I want to print it or use it for something else. Thanks in advance :)

edit

 #include <stdio.h>

 #include <stdlib.h>

 #define MACROs

 #define _CRT_SECURE_NO_WARNINGS



 void xxx(int **a);





 int main(void)

 {

   int *a;



   xxx(&a);



   printf("%d\n\n", a[1]);

 }





 void xxx(int **a)

 {

   int i;



   *a = malloc(5 * sizeof(**a));



   for (i = 0; i < 5; i++)

        a[i] = i;

   printf("%d\n\n", a[1]);

 }

Answer 1

In your main(), you need to have a pointer, not a pointer to pointer. change

  int **a;

to

 int *a;

and, inside the xxx(), change

 a[i] = i;

to

(*a)[i] = i;

That said

Don't use magic numbers, rewrite your malloc statement like
```
*a = malloc(5 * sizeof(**a));
```
to be more robust. Also, for static counts, use #define MACROs.
Please see this discussion on why not to cast the return value of malloc() and family in C..
main() is not a valid signature for a hosted environment. You need to use int main(void), at least.

How to return dynamic array from void function in c?

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