Cannot allocate dynamic array on C - c

Question

Im trying to create a graph structure on C but I got some issues. First, Im getting 2 compilation errors:
main.c:18:19: error: member reference type 'node' is not a
pointer; did you mean to use '.'?
graph[index]->start = NULL;
~~~~~~~~~~~~^~
.
main.c:18:27: error: expression is not assignable
graph[index]->start = NULL;
~~~~~~~~~~~~~~~~~~~ ^
2 errors generated.
compiler exit status 1
I cannot figure out what Im doing wrong. I tried to create an array of nodes* but the compiler doesn't recognize it as a pointer for some reason. It's like malloc doesn't work. Also, I can't manage to acess edge* fields because it's like the array of nodes* is non-existent.
#include <stdio.h>
#include <stdlib.h>
#define maxNodes 4
typedef struct edge {
int target;
struct edge* next;
} edge;
typedef struct {
edge* start;
} node;
void initializeGraph(node* graph) {
graph = (node *) malloc(maxNodes * sizeof(node));
for(int index = 0; index < maxNodes; index++) {
graph[index]->start = NULL;
}
}
int main(void) {
node test;
initializeGraph(&test);
}
Im trying to initialize my structure. Any help is appreciated.

Answer 1

You have a large number of problems in your short example code. As to your error, that is covered by #dbush's answer and [...] serves as a dereference on your pointer making the '.' (dot) operator proper instead of the -> arrow operator.
Next, you cannot declare a node with static storage duration in main() and pass its address for allocation in your function. When you declare node test; all storage is already provided on the stack. You can't then pass that address to your function and allocate additional memory for that struct.
If you intend to have more than one node, then you can either declare an array with static storage duration in main(), or you must declare a pointer in main() and allocate in your function. To make that allocation visible in main(), as noted in my comment, you can either (1) make the return type node * and return a pointer to the allocated block for assignment in the caller, or (2) make the parameter node** and pass the address of your pointer as the parameter.
Putting that altogether and choosing option (1) above, you could do:
#include <stdio.h>
#include <stdlib.h>
#define maxNodes 4
typedef struct edge {
int target;
struct edge* next;
} edge;
typedef struct {
edge* start;
} node;
node *initializeGraph (void) {
node *graph = malloc(maxNodes * sizeof *graph);
if (!graph)
return NULL;
for (int index = 0; index < maxNodes; index++) {
graph[index].start = NULL;
}
return graph;
}
int main (void) {
node *test = initializeGraph();
if (!test)
fputs ("error: initialization failed.\n", stderr);
else
puts ("initialization succeeded");
}
Example Use/Output
$ ./bin/graphinit
initialization succeeded
Allocating For Each test[i].start
Before you can make use of any of the start pointers, you must allocate storage for a struct edge and assign the beginning address for that block of memory to each of your test[i].start pointers. You can do that in your same initializeGraph() function by allocating where you currently set the pointers NULL, e.g.
node *initializeGraph (void)
{
node *graph = malloc(maxNodes * sizeof *graph);
if (!graph)
return NULL;
for (int index = 0; index < maxNodes; index++) {
graph[index].start = malloc (sizeof *graph[index].start);
if (!graph[index].start)
return NULL;
}
return graph;
}
You can then assign a value to the target in each. Extending the earlier example, you could do:
int main (void) {
node *test = initializeGraph();
if (!test)
fputs ("error: initialization failed.\n", stderr);
else
puts ("initialization succeeded");
for (int i = 0; i < maxNodes; i++)
test[i].start->target = i;
puts ("targets filled");
}
Example Use/Output
$ ./bin/graphinit
initialization succeeded
targets filled
(don't forget to free the memory you allocate when it is no longer needed)
Look things over and let me know if you have further questions.

Answer 2

The array index operator [] implicitly dereferences a pointer. The syntax a[b] is exactly the same as *(a + b).
This means that graph[index] has type node, not node *. So use . instead of -> as the error message suggests.
graph[index].start = NULL;