Question

I’m writing a bit of code for a class, but since I have no experience in C I’m a bit unsure of what the code I’ve written actually does. Particularly what the memory looks like. Here’s the relevant bits:

typedef struct listnode *Node;
typedef struct listnode {
    void *data;
    Node next;
    Node previous;
} Listnode;

typedef struct listhead *LIST;
typedef struct listhead {
    int size;
    Node first;
    Node last;
    Node current;
} Listhead;

#define HALLOCSIZE 50

static LIST hallocbuf[HALLOCSIZE];
static LIST *hallocp = hallocbuf;

LIST *CreateList()
{
    if(hallocbuf + HALLOCSIZE - hallocp >= 1)
    {
        LIST temp;
        temp->size = 0;
        temp->first = NULL;
        temp->last = NULL;
        temp->current = NULL;

        *hallocp = temp;
        return hallocp;

    }else
        return NULL;
}

So my question is, in the CreateList function, how is the program allocating memory for temp? And does the code *hallocp = temp copy the temp LIST into the hallocbuf array? I am trying to have all my LIST structs sit in the allocated memory for hallocbuf. Is this what I’m doing? I’m a bit uncertain of how the typedef, structs and pointers play together.

Thanks!

Answer 1

So my question is, in the CreateList function, how is the program allocating memory for temp?

It isn’t, which is a problem. It should do something like this:

temp = malloc(sizeof(Listhead));

And does the code *hallocp = temp copy the temp LIST into the hallocbuf array?

It copies the pointer that was saved in temp into the first element of hallocbuf (assuming that hallocp hasn’t been changed anywhere and still has the value that it has been initialized to, pointing to hallocbuf[0]).

Generally it’s not usually a good idea to hide the fact that LIST and Node are pointers behind typedefs. It’s much clearer where memory needs to be allocated of freed if it’s obvious which variables are pointer, and having an explicit * in the variable declaration makes that clear.