Is this creating a linked list?

I typedef'd some stuff:

Code:

typedef struct country Country;

struct country {
  char name[LENGTH];
  int dimen;
  int num;
  //QTnode *flag;
  Country *next;
};

And here is the function that makes the node:

Code:

 Country *makeNode(char p[MAXLENGTH]){

        int i = 1;


        Country *node;
        node = (Country*)malloc(sizeof(Country));
        strcpy(node->name, p);
        getImage(node->name, &node->dimen);
        node-> num = i;
        node-> next = NULL;
        printf("*%d [%d] %s\n", node->num, node->dimen, node->name);
        return (node);
}

Suppose the user presses "A", and this function gets activated and the following commands are executed properly. If the user presses "A" again, is this node destroyed and a new one created? If so, how should I go about retaining this node and creating a new node and having a pointer to the new node so that new information is stored there and can be accessed later?

Cheers

Since makeNode allocates the memory necessary to make an instance of Country, a new node is created in that function and returned to the caller. Nothing is destroyed nor will it ever be until the node is unlinked from the data structure and explicitly freed.

You can access a node again later by linking it in the data structure - where the node is placed in the data structure depends entirely how you assign the return value.

So currently if two nodes are made by my function, they aren't linked?

Nope.

How would I link them?

I told you. Assign the return value. In a singly linked list this is especially simple.

Code:

Country *first    =   makeNode( "America" );
first->next       =   makeNode( "France" );
first->next->next =   makeNode( "Germany" );

You now have a singly-linked list with a NULL tail. You're free to allocate an extra dummy node to serve as a tail as well, rather than relying on a simple NULL comparison.

This idea is otherwise known as insertion. That is commonly implemented as a function that attaches a node parameter to another list, creating a larger list.

I think I see what you mean, thanks for the pointers, I'll see if I can implement them now.

Cheers