Lecture 19
Put your content here . . .
three ways to allocate memory:
1. static---->know the size when writing the program
eg. int arr[100];
vector v[200];
2. know the size somewhere during execution
eg. int length;
scanf("%d", &length);
int *arr;
arr = malloc(sizeof(int)*length);
3.grow and shrink based on ran-time needs(dynamic structures)
eg. linked list, binary tree(list mode, tree mode)
type of struct dstructure { int value; vector vec; Person *p;
struct dstructure *next;(linked list) struct dstructure * left;(binary tree) struct dstructure * right;(binary tree) } Node
streaming data
Node *n; n = malloc(sizeof(Node));
typeof struct listnode { int value; struct listnode *next; }Node;<---(don't forget)
(next few lectures, focus on linklist only, and data type would just be integer)
Node *n= NULL; n = malloc(sizeof(Node)); n->value = 264; Node *n2; n2 = malloc(sizeof(Node)); n2->value = 2012; n2->next = NULL; n->next = n2;
(abstract view below) assign one's value to another's link to link them together
Node *n3; n3 = n; printf("%d",n3->value);(=264) n3 = n3->next; printf("%d", n3->value);(=2012)
Node *Node.construct(int v) { Node *n; n = mallloc(sizeof(Node)); n->value = v; n->next = NULL;(<----don't forget) return n; }
Node *List_insert(Node *t, int v) { Node *n = Node.construct(v); n->next = t; return n; }
Node *head = NULL; head = List.insert(head, 264); head = List.insert(head, 2012);
Node *head = NULL; head = List.insert(head, 264); head = List.insert(head, 2012);
Class Notes: James Chen
three ways to allocate memory
1. -> know the size when writing the program;
int arr[size];
vector v[size];
2. -> know the size somewhere during execution
int length;
scanf("%d", &length);
int *arr;
arr = malloc(sizeof(int)*length);
3. -> grow and shrink -- dynamic structure
link listed
binary tree
typedef struct dstructure //treenode, listnode { /*data*/
int value; Vector vect; Person *p
/*link*/
/*link list*/ struct dstructure *next; //same name as in header (dstructure) /*binary tree*/ struct dstructure *left; //see dstructure struct dstructure *right;//see dstructure
}Node;
// name of structure is Node, dstructure relates to its organization, name is assigned links are pointers, needed to store address because the structure has not ended yet can link the list so only one value is needed to store the entire array
//how to create linked nodes: //create original node
n = malloc(sizeof(Node)); n->value = 264; n->next = NULL;
//create new node, link to existing node
Node *n2; n2 = malloc(sizeof(Node)); n2->value = 2012; n2->next = NULL; n->next = n2;
//constructor example Node *Node_construct(int v) {
Node *n; n = malloc(sizeof(Node)); n->value = v; n->next = NULL; return n;
} Node *List_insert(Node *insert, int v) //starting (original) node {
Node *n = Node_construct(v); n->next = head; return n;
}// program creates from the end of the list
int main() {
head=List_insert(head,264); head=List_insert(head,2012); return(0);
}
