C中的动态大小数组

Question

I am trying to make use of Array of nodes with dynamic memory allocation so that I can increase the number of nodes at runtime. However I seem to be getting an error which is unknown to me. I am probably using the array in wrong way, so please check and correct me. listNo used in the code is an integer variable.

Code:

Node* lists = (Node*) malloc(100 * sizeof(lists));

printf("\n Enter the number of lists:");
scanf("%d", &nbrOfLists);

if(nbrOfLists < 0)
    return -1;
if(nbrOfLists>100)
    lists = realloc(lists, 100 * sizeof(lists));


lists[listNo] = NULL;  // getting error here incompatible types assigning Node from type 'void*'
lists[listNo]= insertValue(lists[listNo], val); 

I mean each element in an array has first element of an individual linked list. The next element is another independent first node of another linked list.

Answer 1

For linked list you do not need to allocate memory for all elements. You have to have single root element Node * root and link all other to it. No need from malloc and realloc - that is the idea of "linked list". In your example you have dynamic array of objects.

Answer 2

1. You should learn pointers concepts. Code signals you don't understand it well. Just friendly long term note.
2. lists has type of pointer to Node which is "address", in 32-bit OS it is 4 bytes, in 64 bit - 8 bytes. But no relation to actual Node structure size.
3. When you do `malloc(N * sizeof(lists)) you actually allocate N elements, each of which is address (4 or 8 bytes, again no relation to actually stored information).
4. When you refer to lists[X] you actually say "take this address of Node , and then take X-th Node under it". You mean Node , not "address of Node " so your compiler says "I cannot assign Node with NULL".

Hope this helps. Please re-check pointers concept in C. Blocking issue for many people to grow further.

To say "please allocate N elements of type Node " you should write something like:

Node *lists = realloc(lists, N * sizeof(*lists));

And you'd better say Node *list than Node* list (practically it results the same code but please check what is the difference - it is basic principle and you should learn it yourself).

Answer 3

If you want an array of pointers to nodes, then you need a pointer to the first pointer to node of an array of pointers to nodes. This would be a double pointer for a variable sized array: node **arrayofpoitnerstonode; , then arrayofpoitnerstonode[0] would be the first pointer to a node, arrayofpoitnerstonode[1] would be the second pointer to a node, ... .