Basically, array should be initialized on user input. If input = 3, that means this array can store one linked list in index 0,1,2 respectively (so total 3 lists)
int input = 3;
list* array[n]//not allowed as n is not constant, also not in heap so can't pass it across functions
list* array[] = (list*) malloc(sizeof(list)*input)//compiler error
Preparing for interviews...so you can say home-work!
An array of linked lists could be either an array of head nodes (assuming a standard linked list implementation) or an array of pointers to lists. In either case, the problem you seem to be facing is how to dynamically allocate an array.
The generic syntax for dynamically allocating an array on the heap is
type * array = calloc(number_of_elements, sizeof(type))
Correcting the compilation error in your code would thus be
int input = 3;
list ** array = calloc(input, sizeof(list*));
Is this what you are looking for?
It should be list* array = malloc(sizeof(list) * input), malloc returns base address of the newly allocated memory location. You can use this as the array i.e. you can access array[0]..array[input - 1].
A (single) linked list is often a structure with a pointer to the next structure. This pattern makes it easy to add, delete and insert things from that list and still keeps the whole data usage flexible and governed in run-time.
a linked list in your situation will look like this
struct List
{
// contents of the list
List* Pointer_to_next_list;
};
Then you can add functions for keeping track of each list. I suggest reading Wikipedia::Linked List for how this works.
