Getting value from struct by pointer to pointer in C

Question

I'm making a program in C. Basically I have a struct that contains some fields, and an array of pointers. Each struct has an pointer array that points to another struct forming a "connection" between them. I'm trying to get the value of a field that is stored at the memory address the pointer that is pointing at it.

Suppose this. I have two of these nodes in memory. A and B. A has a pointer inside A's array that is a reference to B. To get this pointer I'd have to do something like this:

*temp_ptr = (*ptr).pointer_array[0]

This would get the pointer address *ptr and give it to *temp_ptr.

Now what I am wondering is this. How can i do this? When I try this, I get "Expression must have struct or union type"

When I try this in lets say Java I could do this

int variable = field[0].fieldIWantToGet

I'd get the desired outcome.

Heres an image to clarify the intended behaviour that I'm trying to get. Link to behavior

Where Struct A is in a "global" collection of structs and has a array of pointers that lead to other Structs, such as B

Here is some code from my project.

#define GLOBAL_PTR_ARRAY_SIZE 10

Node* global_node_array[10];  

typedef struct Node{

unsigned char node_id;
int *ptr_array[10];
int ptr_array_size;


}Node;



void append_connection(short position, short destination) {   

    Node* position_ptr = global_node_array[position];        
    Node* destination_ptr = global_node_array[destination];

    if ((*position_ptr).ptr_array_size < GLOBAL_PTR_ARRAY_SIZE) {   
        int current_ptr_array_size = (*position_ptr).ptr_array_size;  

        (*position_ptr).ptr_array[current_ptr_array_size] = destination_ptr;

        (*position_ptr).ptr_array_size++;

}

void print_id(Node* ptr) {

    node* dptr = NULL;

    dptr = ptr->ptr_array[0];

    pptr = (int) (*ptr).ptr_array[0];

    fprintf(stdout, "%d connection to %d exists", (*ptr).node_id, dptr- 
   >node_id);      
}


int main(int argc, char const *argv[])
{
append_connection(0,1);
print_id(global_node_array[0]);

return 0;
}

(*ptr).pointer_array[0] can be simplified to ptr->pointer_array[0] — lost_in_the_source
– lost_in_the_source, Commented Feb 24, 2019 at 18:36
"This would get the pointer address *ptr and give it to *temp_ptr." Why do you think this? There is more on the right-hand side than just *ptr, and you've ignored those extra bits. Please present your minimal reproducible example. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Feb 24, 2019 at 18:38

0___________ · Accepted Answer · 2019-02-24 18:46:47Z

2

Your picture shows array of the structs not the pointers. But the example below covers both.

struct a{
    int field1,field2;
}

struct b{
    struct a m[10];
}

struct c{
    struct a *m[10] 
}




   /* and usage */        
    struct c *x;
    struct b *y;

    x -> m[5].field1;
    y -> m[5] -> fileld1;

answered Feb 24, 2019 at 18:46

0___________

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