need help in understanding passing char* [] to function - c

Question

I am trying to pass a array of pointers to string to a function where I need to set the values. In the passing function I do not know the number of strings I will get, the called function is calling some other function which returns list of strings.
Sample code below:
int main() {
char** list;
create(list);
}
int create(char **array) {
char* str[] = { "hello", "dear" };
int len;
int i = 0;
for (i = 0; i < 2; i++) {
len = strlen(str[i]);
printf("%d\n", len);
*(array + i) = (char*) malloc(len * sizeof(char*));
strcpy(*(array + i), str[i]);
i++;
}
return 1;
}
This gives me segmentation fault.
What wrong am I doing here. Please help.
Thanks
EDIT
Updated code from below comments:
int main() {
char** list;
create(list);
int i = 0;
for (i = 0; i < 2; i++) {
printf("%s\n", list[i]); // segmentation fault
}
}
int create(char **array) {
char* str[] = { "hello", "dear" };
int len;
int i = 0;
array = malloc(2 * sizeof(char*));
for (i = 0; i < 2; i++) {
len = strlen(str[i]);
printf("%d\n", len);
*(array + i) = (char*) malloc(len * sizeof(char));
strcpy(*(array + i), str[i]);
printf("%s\n", array[i]); // this prints
}
return 1;
}
Now getting segmentation fault in main while printing the list.
Actual code where I am reading the strings
int i;
for ( i=0; i<reply->elements; i++ )
{
printf( "Result: %d---%s\n", i,reply->element[i]->str );
*array[i] = (char*)malloc(strlen(reply->element[i]->str));
printf("***");
strcpy(array[i],reply->element[i]->str);
printf( "Array[%d]: %s\n", i,array[i] );
}

Answer 1

You correctly alloc memory for the individual strings, but fail to alloc some for the array itself.
You should use:
int main() {
char* list[8] = {0}; /* initialize pointers to NULL */
create(list);
/* free allocated memory - free(NULL) is legal and is a noop */
for (i=0; i<sizeof(list)/sizeof(list[0]); i++) free(list[i]);
return 0; /* never return random value from main */
}
And you should remove the i++ at the end of the loop in function create because it leads to a double increment.
Alternatively you could alloc the array itself in the function create:
int create(char ***array) {
char* str[] = { "hello", "dear" };
int len;
int i = 0;
*array = malloc(1 + sizeof(str)/sizeof(str[0]));
for (i = 0; i < 2; i++) {
len = strlen(str[i]) + 1;
printf("%d\n", len);
(*array)[i] = malloc(len * sizeof(char*));
strcpy((*array)[i], str[i]);
}
(*array)[i] = NULL;
return i;
}
int main() {
char** list;
create(&list);
}
In above code, the length of the array is the return value from create, and the last element of list is a NULL (in the same logic as argc/argv).

Answer 2

You need to allocate some space for list or undefined behavior occurs:
char* list[2];
You increment i twice; therefore, remove the i++ from the bottom of the for loop.
Minor notes:
refer to string literals as const char*
use array[i] instead of *(array + i)
don't cast the result of malloc
malloc allocates too much space as you allocate len char*s, even though you need just chars. Also, as #CoolGuy noted, you need one extra byte for the null byte. Replace the allocation with
array[i] = malloc(len * sizeof(char) + sizeof(char));
or
array[i] = malloc(len + 1);
call free after malloc
you assign 0 twice to i; remove the initialization

Answer 3

You allocate two arrays (char*) to store the strings "hello" and "dear" but does not allocate the array (char**) containing those two string array.

Answer 4

I would suggest you to change declaration of function create to this -
int create(char ***array);
And call it like this -
create(&list);
In function create allocate memory like this -
*array = malloc(2 * sizeof(char*));
for (i = 0; i < 2; i++)
{
len = strlen(str[i]);
printf("%d\n", len);
(*array)[i] =malloc(len * sizeof(char*)+1);
strcpy((*array)[i], str[i]);
}
And do the printing as you do in main.
Note - free memory that you allocate.
And you should declare len as type size_t -> size_t len;
and print it with %zu specifier in printf .
See working code here -https://ideone.com/GX2k9T