pointers (additional material)
DESCRIPTION
Pointers (additional material). Array of pointers. Problem : Write program that receive strings from a user and print these strings in a lexicography order. Solution : Array of pointers to char (malloc) char * names[256] or char ** names; This is not the same declaration !!. - PowerPoint PPT PresentationTRANSCRIPT
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Pointers(additional material)
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Array of pointers
Problem : Write program that receive strings from a user
and print these strings in a lexicography order.
Solution :Array of pointers to char (malloc)
char * names[256] or char ** names;
This is not the same declaration !!
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Array of pointers
Declaration:char ** names;int n,i;scanf(“%d”,&n);names = (char **)malloc(n * sizeof(char *));for(i=o; i<n; i++){
char[i] = (char*)malloc(256*sizeof(char));
}// bubble on the pointers !!
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Complete problem
We use exactly the size for the input strings.
Assuming that the strings have at most 255 letters.
We need receive a sorted array of strings from the function.
All the inputs are from the user.
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Complete solution
char ** sorted_list(){char ** names, temp[256], * temp2;int n, i, flag;scanf(“%d”,&n);names = (char **)malloc(n * sizeof(char *));for(i=0; i<n; i++){
gets(temp);char[i] = (char*)malloc(strlen(temp)+1);strcpy(char[i],temp); // char + i
}/* Sorting pointers by lexicography string
order */
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Complete solution(cont.)
// sorting do{
flag =0;for(i=0 ; i< n-1 ; i++)
// names + i , names +i+1if( strcmp(names[i],names[i+1]) >0){temp = names+i;names +i = names +i+1;names+i+1 = temp;flag = 1;}
}while(flag);return names;
}
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Linked Lists
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Department of Computer Science-BGU
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Problems with dynamic arrays
Problems: Adding/ delete member. Reallocation. Building sort list . Merging .
Solution: Linked list Simple add/delete member. No need reallocation. Building sorting. Simple merging.
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Linked lists?
head
NULL
Structures
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Department of Computer Science-BGU
A better alternative might be using a linked list, by “self reference”.
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Linking students
typedef struct Student_t {char ID[ID_LENGTH];char Name[NAME_LENGTH];int grade;struct Student_t *next; /* A pointer to the next item on the list */
} item;
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Different definition
A different use of typedef:
typedef struct Student_t item ;struct Student_t{
char ID[ID_LENGTH];char Name[NAME_LENGTH];int grade;item *next; /* A pointer to the next item
on the list without use of “struct” in the pointer definition */
} ;
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Creating a new kind of student
Usually when using linked lists we don’t know how many elements will be in the list
Therefore we would like to be able to dynamically allocate new elements when the need arises.
A possible implementation follows…
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Creating a new kind of student
item*create_student(char * name, char * ID, int grade) {item *std;std = (item *)malloc(item));if (std == NULL) {
printf(“Memory allocation error!\n”);exit(1);
}strcpy(std->Name, name);strcpy(std->ID, ID);std->grade = grade;std->next = NULL;return std;
}
std
NULL
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Linked lists - insertion
…
Head
Insert new item:
Previous
Next
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Linked lists - insertion
…
Head
Insert new item:
Previous
Next
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Linked lists - insertion
…
Head
Insert new item:
Previous
Next
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Adds a new item to the end of the list
head NULL
newItem
NULL
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Adds a new item to the end of the list
newItem
NULLhead
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Adds a new item to the end of the list
NULL
head
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Adds a new item to the end of the list
head
NULL
newItem
NULL
currItem
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Adds a new item to the end of the list
head
NULL
newItem
NULL
currItem
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Adds a new item to the end of the list
head
NULL
newItem
NULL
currItem
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Adds a new item to the end of the list
head
NULL
newItem
NULL
currItem
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Adds a new item to the end of the list
head
NULL
newItem
NULL
currItem
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Adds a new item to the end of the list
item * add_last(item *head, item* newItem){item *currItem;if (!head)
return newItem;currItem = head;while(currItem->next)
currItem = currItem->next;currItem->next = newItem;
return head;}
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Inserts into a sorted list
head NULL
newItem
NULL
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Inserts into a sorted list
newItem
NULLhead
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Inserts into a sorted list
NULL
head
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Inserts into a sorted list
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newItem
NULL
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Inserts into a sorted list
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Inserts into a sorted list
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Inserts into a sorted list
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Inserts into a sorted list
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currItem
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Inserts into a sorted list
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Inserts into a sorted list
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Inserts into a sorted list
// while keeping it sorted ascending by keyitem * insert(item *head, item *newNode) { item *currItem; if (!head) return newNode; //check if newNode's key is smaller than all keys and should be first
if (newNode->key < head->key) { newNode->next = head; return newNode; } currItem = head; while (currItem->next && newNode->key > currItem->next->key)
currItem = currItem->next;//put newNode between currItem and currItem->next //(if currItem is last then currItem->next == NULL) newNode->next = currItem->next; currItem->next = newNode; return head;}
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Linked lists - searching
…
head?currItem
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Linked lists - searching
…
head?currItem
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Linked lists - searching
…
head?currItem
!
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Searching for an item
//searches for an item with passed key.//returns NULL if didn't find it.item *search(item *head, int key) {
item *currItem = head; if (!head) return NULL;while (currItem) { //loop through the list
if (currItem->key == key)return currItem;
currItem = currItem->next;} //didn't find the item with the requested keyreturn NULL;
}
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Department of Computer Science-BGU
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Print list’s members
//prints keys of items of the list, key after key.void printKeys(item *head) {
item *curr = head; while (curr) {
printf("%d ", curr->key);curr = curr->next;
}putchar('\n');
}
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Linked lists - delete
…
Head
Structure to delete
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Linked lists - delete
…
Head
Current
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Previous
NULL
Structure to delete
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Linked lists - delete
…
HeadPreviou
sCurrent
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Structure to delete
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Linked lists - delete
…
HeadPreviou
sCurrent
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Structure to delete
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Linked lists - delete
…
HeadPreviou
sCurrent
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Linked lists - delete
…
HeadPreviou
sCurrent
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Linked lists - delete
…
HeadPreviou
s
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Remove the item with a given value
item *remove(int value, item* head){item * curr= head,*prev=NULL;int found=0;if(!head)
printf("The LL is empty\n");else{
while(curr)if(value==curr->value){
prev ?prev->next=curr->next:head=head->next;free(curr);found=1;break;
}else{
prev=curr;curr=curr->next;
}if(!found) printf("The record with key %d was not found\
n",value);}
return head;}
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Freeing students
After we’re done, we need to free the memory we’ve allocated
One implementation is as we saw in class
void free_list(Student *head){
Student *to_free = head;
while (to_free != NULL) {
head = head->next;free(to_free);to_free = head;
}}
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Recursive freeing
A perhaps simpler way to free a list is recursively.
void free_list(Student *head){
if (head== NULL) /* Finished freeing. Empty list */return;
free_list(head->next); /* Recursively free what’s ahead */free(head);
}
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Split linked list
typedef struct stam item;struct stam{
int value;item * next;
}void main(){
item * head=*odd=*even = NULL;//// /* build list */// Now split it in two lists with odd and even membersvoid Split(head, odd, even);print_list(odd);print_list(even);
}
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Split linked list
// Split the nodes to these 'a' and 'b' listsvoid Split(item * source, item * od, item * ev) {
item * current = source;while (current != NULL) {
temp = current;if(temp->value %2 != 0)od = add_last(od,current);else ev = add_last(ev,current);current = current -> next;temp->next = NULL;
}}
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Split linked list
typedef struct stam item;struct stam{
int value;item * next;
}void main(){
item * head=*odd=*even = NULL;//// /* build list */// Now split it in two lists with odd and even membersvoid Split(head, &odd, &even);print_list(odd);print_list(even);
}
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Split linked list
void Split(item * source, item ** od, item ** ev) {item * current = source;item * a =*b = NULL;while (current != NULL) {
temp = current;if(temp->value %2 != 0)
a = add_last(a,current);else b = add_last(b,current);current = current -> next;temp->next = NULL;
}*od = a;*ev = b;
}
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Split linked list
// Split the nodes to these 'a' and 'b' listsvoid Split(item * source, item ** od , item ** ev) {
item * a = NULL; item * b = NULL;item * current = source;while (current != NULL) {
MoveItem(&a, ¤t); // Move a node to 'a'if (current != NULL) { MoveItem(&b, ¤t); // Move a node to 'b'{
{*od= a;*ev = b;
{
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