You’ve got several options:
1) Allocate your array on the heap using malloc()
, and return a pointer to it. You’ll also need to keep track of the length yourself:
void give_me_some_chars(char **arr, size_t *arr_len)
{
/* This function knows the array will be of length 2 */
char *result = malloc(2);
if (result) {
result[0] = 'c';
result[1] = 'a';
}
/* Set output parameters */
*arr = result;
*arr_len = 2;
}
void test(void)
{
char *ar;
size_t ar_len;
int i;
give_me_some_chars(&ar, &ar_len);
if (ar) {
printf("Array:\n");
for (i=0; i<ar_len; i++) {
printf(" [%d] = %c\n", i, ar[i]);
}
free(ar);
}
}
2) Allocate space for the array on the stack of the caller, and let the called function populate it:
#define ARRAY_LEN(x) (sizeof(x) / sizeof(x[0]))
/* Returns the number of items populated, or -1 if not enough space */
int give_me_some_chars(char *arr, int arr_len)
{
if (arr_len < 2)
return -1;
arr[0] = 'c';
arr[1] = 'a';
return 2;
}
void test(void)
{
char ar[2];
int num_items;
num_items = give_me_some_chars(ar, ARRAY_LEN(ar));
printf("Array:\n");
for (i=0; i<num_items; i++) {
printf(" [%d] = %c\n", i, ar[i]);
}
}
DO NOT TRY TO DO THIS
char* bad_bad_bad_bad(void)
{
char result[2]; /* This is allocated on the stack of this function
and is no longer valid after this function returns */
result[0] = 'c';
result[1] = 'a';
return result; /* BAD! */
}
void test(void)
{
char *arr = bad_bad_bad_bad();
/* arr is an invalid pointer! */
}