How to do linear search using pointers? - c++

Question

I'm still new to programming. I just wanted to ask how do I to do linear search using pointers. I wanted to make a book management program and i have made a program with pointers written.
This is the example of how i want it.
This is the coding
#include <iostream>
#define MAX 5
using namespace std;
struct record
{
int id;//stores id
float price;//store price
int qty;//stores quantity
record* next;//reference to the next node
};
record* head;//create empty record
record* tail;//the end of the record
void push(record *& head, record *&tail, int id, float price, int qty)
{
if (head == NULL)
{
record* r = new record;
r->id = id;
r->price = price;
r->qty = qty;
r->next = NULL;//end of the list
head = r;
tail = r;
}
else if (head != NULL && (MAX - 1))
{
record* r = new record;
r->id = id;
r->price = price;
r->qty = qty;
r->next = head;
head = r;
}
}
int pop(record *&head, record *& tail)
{
if (head == NULL)
{
cout << "No record in memory" << endl;
}
else if (head == tail)
{
cout << "The record "<<"ID: " << head->id << "\nPrice: " << head->price << "\nQuantity: " << head->qty << "\n" << "was deleted" << endl; //CORRECTION HERE
}
else
{
record* delptr = new record;
delptr = head;
head = head->next;
cout << "The record " << delptr->id << ", " << delptr->price << ", " << delptr->qty << " was deleted" << endl; //CORRECTION HERE
delete delptr;
}
return 0;
}
void display(record *&head)
{
record* temp = new record; //CORRECTION HERE
temp = head;
if (temp == NULL)
{
cout << "No record in memory" << endl;
}
else
{
cout << "Record : " << endl;
while (temp != NULL)
{
cout <<"\nID: "<< temp->id << "\nPrice: " << temp->price << "\nQuantity: " << temp->qty <<"\n"<< endl; //CORRECTION HERE
temp = temp->next;
}
}
}
int LinearSearch(record *&head) {
}
char menu()
{
char choice;
cout << "\t::MENU::\n" << endl;
cout << "1. Add new record\n" << endl;
cout << "2. Delete record\n" << endl;
cout << "3. Show record\n" << endl;
cout << "4. Quit\n" << endl;
cout << "-----------------------\n" << endl;
cout << "\nEnter selection : " << endl;
cin >> choice;
return choice;
}
int main()
{
record* head;
record* tail;
head = NULL;
tail = NULL;
char choice;
do
{
cout << "---------------------- - \n" << endl;
choice = menu();
switch (choice) { //CORRECTION HERE
case '1':
int id, qty;
float price;
cout << "Enter ID:";
cin >> id; // Please correct yourself here, what is r here, r is not declared anywhere
cout << "\nEnter Price: ";
cin >> price;
cout << "\nEnter Quantity: ";
cin >> qty;
push(head, tail, id, price, qty);
break;
case '2':
pop(head, tail);
break;
case'3':
display(head);
break;
default:
cout << "Quiting...\n";
}
} while (choice != '4');
return 0;
}
How do I write linear search of pointer code for this coding? I tried finding examples throughout the web and when i execute it, it didn't work so i just leave it blank.

Answer 1

Well, I saw you have a list and you are dealing with pointers.
If you want to do linear search in a record id, for example, you can do it like this:
record *aux = head;
while(aux != NULL){
if(aux->id == id_you_want_to_find){
printf("I found it\n");
}
aux = aux->next;
}
you usually use object.attribute to access the attributes of common objects, but when you have a pointer to an object, you must do pointerToObject->attribute.

Answer 2

You can write one if you want but there is no need to when there already exists libraries that do this for you. Since you are using a list structure I show this using a simple std::list. You could also change this to a std::vector and just do a simple for loop iteration using index notation since the speed of search through them is constant as opposed to linear. Here is one method of doing a search through a list linear.
#include <list>
record* searchRecords( std::list<record>& records, int id ) {
if ( records.empty() ) {
std::ostringstream strStream;
strStream << __FUNCTION__ << " Invalid list of records: list is empty.";
throw ExceptionHandler( strStream ); // Not Written, but what should be done instead of returning.
return nullptr;
}
std::list<record>::iterator it = records.begin();
while ( it != records.end() ) {
if ( it->id == id ) {
return (&(*it));
}
++it;
}
std::ostringstream strStream;
strStream << __FUNCTION__ << " No entry found in search with ID{" << id << "}.";
Logger::log( strStream, Logger::LOGGER_INFO ); // Not implemented here same as above for ExceptionHandler
return nullptr;
}
Since linked lists are not associative they Must be traversed from beginning to end for every entry N in the list to either insert, find or delete. The time complexity here is linear.
If you want faster time insertion time with large lists you can use <multiset> if there may be duplicate items or <set> if every known item is unique. These have instant insertion. If you want constant search and don't care about insertion time then <vector> is what you would want.

Answer 3

The normal answer would be: don't write a linear search yourself, it's called std::find_if. However, C++ expects that your datastructure exposes iterators. An iterator refers to a record (or the end of your list). You get the actual record by calling operator* on the record, and you get the next record by calling operator++.
Yes, this is similar to pointers. That's intentional; pointers are iterators for contiguous arrays. That means you can call std::find_if on an array. But since you chose to implement a linked list instead of an array, you'd' need to implement your own iterator class.