需要一些帮助来理解C中的指针和内存

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.

Answer 2

temp is allocated in the space used for objects with "automatic storage duration" - this is usually on a runtime stack, but you don't really need to know the details. The space is deallocated when the block in which it was allocated is exited (in your case, when you hit the return ).

The line *hallocp = temp; does indeed copy the value of temp into the memory that hallocp is pointing at, which is hallocbuf[0] .

The problem is that temp is just a pointer itself - and it's not pointing at anything. This is called a "dangling pointer". This means that when you try to access what it's pointing at, you have an error. That happens in these lines:

temp->size = 0;
temp->first = NULL;
temp->last = NULL;
temp->current = NULL;

You can't have your structs sit in the memory allocated for hallocbuf , because it doesn't have room for structs - it's just an array of pointers, not an array of structs.

Answer 3

If LIST were

typedef struct listhead LIST;

and you accessed temp

temp.size = 0;
...

then

*hallocp++ = temp;

would use hallocp as a pointer into the hallocbuf buffer and place the newly initialized element there. Not the best way to do it, though. You could replace temp with

hallocp->size = 0;
...
++hallocp;