Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

how do you search for a lastname in a linked list? c++

Computer Science
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
i have a linked list that stores firstname, lastname, and phone number. i need to find the lastname
Are you using the STL?
I am using visual studio c++

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Did you implement the linked list yourself?
i did. ammm can you give me an example of a referenced variable?
Referenced variable is like a pointer that is pretending to be deferenced.
Okay I need to know what your code is like first of all.
bool PhoneBook::Lookup (const string& lname, const string& fname, string& pnum) const { PhoneBookItem* p = head; PhoneBookItem* q = head->next; if (head == NULL) return false; else if (lname == p->lastname) { if (fname == p->firstname) { return true; } else return false; } else { while (q != NULL && lname != q->lastname) { p = q; } if (q != NULL && lname == q->lastname) if (fname == q->firstname) return true; else return false; else return false; } }
It says that the pnum should be returned to the calling function via a third argument, a reference variable. I don't know how. can you help me with this?
What is pnum supposed to be?
wio..phone number
All you need to do is assign pnum to the pnum of the node with the correct name.
``` while (q != NULL && lname != q->lastname) { p = q; } ``` This code doesn't make sense. You should try something like: ``` while (q != NULL && lname != q->lastname) { p = q; q = q->next; } ``` You need to update q and p.
So assuming you find it, before you return true, you wanna assign pnum. ``` if (fname == p->firstname) { pnum = q->pnum; // put it here return true; } ``` ``` if (fname == q->firstname) { pnum = q->pnum; // here too return true; } else ```
I tried that, but it doesn't work. Here is the code in the main. void LookupEntry(const PhoneBook& pb) { string last, first, phone; cout << "Enter last name: "; cin >> last; cout << "Enter first name: "; cin >> first; bool success = pb.Lookup(last, first, phone); if (success) { cout << "\nThe phone number is " << phone << ". \n"; } else { cout << "\nError: no one by this name is in the phone book.\n"; } }
``` string last, first, phone; ``` The problem is that `phone` should be a reference. ``` string last, first; string& phone; ```
Does it even compile?
It does, but that code in the main cannot be changed since our professor said so.
Okay, what sort of output are you getting? What is happening?
It outputs : Error: no one by this name is in the phone book. ================================================= if (success) { cout << "\nThe phone number is " << phone << ". \n"; } else { cout << "\nError: no one by this name is in the phone book.\n"; }
Alright, are you searching for something that is in the list?
yes
``` else if (lname == p->lastname) { if (fname == p->firstname) { return true; } else return false; } ``` This code is wrong. If the last name is correct but the first name is not correct, it should continue searching, but your code instead gives up entirely.
``` else if (lname == p->lastname && fname == p->firstname) { return true; } ``` This is what you should be doing.
There are a lot of bugs in your code, it'd be easier to show the correct way...
For example, why have `p` and `q`? You only need one to keep track of where you are, right?
yeah but i need to test this ============================================= if (head == NULL) return false; else if (lname == p->lastname && fname == p->firstname) { p->phone = head->phone; return true; } else { while (q != NULL && lname != q->lastname) { p = q; p = p->next; } if (q != NULL && lname == q->lastname && fname == q->firstname) { q->phone = head->phone; return true; } else return false; } ===================================== else { while (q != NULL && lname != q->lastname) { p = q; p = p->next; } if (q != NULL && lname == q->lastname && fname == q->firstname) { q->phone = head->phone; return true; } else return false; }
why are you setting `p->phone` to `head->phone`?
those are the variable in my struct ====================================================== struct PhoneBookItem { string lastname; string firstname; string phone; PhoneBookItem* next; PhoneBookItem(const string& l, const string& f, const string& ph); };
Yeah, but you're setting the current item `p` to have the same phone number as the first item `head`. There is no reason to do that.
Okay, you should start out with your current item point to the front: ``` PhoneBookItem* current = head; ```
yes i did that
If you go through the whole list without finding it, then you want to return false. ``` while (current != NULL) { // search code here current = current->next } return false; ```
Does that loop make sense so far? @Lynncake
yes it does, i'm writing the rest of it. but i am not sure if this is right
PhoneBookItem* p = head; if (head == NULL) return false; else { while (p != NULL) { if (lname == p->lastname && fname == p->firstname) { return true; } p = p->next; } return false; }
Now, for the search code... we just check if the names are correct ``` if (current->firstname == fname && current->lastname == lname) { pnum = current->phone; return true; } ```
We don't have anything if the names don't match. The loop will take care of that for us.
yes it worked! thank you
One second though.
``` if (head == NULL) return false; else { while (p != NULL) ``` The `if (head == NULL)` here is pointless. The `while (p != NULL)` already check if `head` is `NULL`.
@Lynncake Does that make sense?
yes, it does thanks a lot
alright, well good luck then.
for the delete, is this right? ====================================== PhoneBookItem* p = head; while (p != NULL) { if (lname == p->lastname && fname == p->firstname) { delete p; p->next=NULL; num--; return true; } p = p->next; } return false; ====================================== I followed your method/concept.
The problem with this is that you need to keep track of the item before you.
And you set the guy behind you's `next` property to your own `next` property.
so in here i should hava another pointer that points to the next element?
This case is a bit tricky because you have two case: 1) They are deleting the head... 2) They are deleting something in the middle of the list
For 1) you delete it, and set head to NULL For 2) you need to set the previous guy's next to the deleted guy's next
``` PhoneBookItem* last = head; PhoneBookItem* current = head->next; // Check head // Check guys in the middle of the list ```
For check head, we wanna do: ``` if (head != NULL && head->firstname == fname && head->lastname == lname) { delete head; head = NULL; } ```
For checking others, we want a loop: ``` while (current != NULL) { \\ check current last = current; current = current->next; } return false; ```
For current, we do something similar to head, but with a change: ``` if (current->firstname == fname && current->lastname == lname) { last->next = current->next; delete head; return true; } ```
Hmmm, there should be a `return true;` in the check head part too.
What we are doing is changing this: |dw:1364256150764:dw|
so the last is the head?
To this: |dw:1364256188626:dw|
`last` starts out as the head in the beginning. But last is set to `current` every time go move to the next element. `last` is just storing what comes before `current`
so i should create another pointer pointing to last?
Yep
I gotta brb for a bit, but look at the code I posted if you are confused
i will, and thank you so much
yes it did delte, but now my printing is messed up after deleting ============================================== ///////////////////////DELETE//////////////////////////////////////// PhoneBookItem* p = head; PhoneBookItem* q; int count=0; while (p != NULL) { if (count==0) { if (p->firstname == fname && p->lastname == lname) { delete head; head = NULL; num--; return true; } } else if (count>0) { if (p->firstname == fname && p->lastname == lname) { q->next = p->next; delete head; num--; return true; } } q = p; p = p->next; count++; } return false; ============================================= ///////////////////////////////PRINT///////////////////////////////// ostream& operator << (ostream& out, const PhoneBook& pbook) { PhoneBookItem* p = pbook.head; while (p != NULL) { out <lastname<<", "<firstname<<": "<phone<next; } return out; }
There is nothing wrong with printing. The linked list is getting ruined by your delete somehow.
``` if (p->firstname == fname && p->lastname == lname) { q->next = p->next; delete head; num--; return true; } ``` You should be deleting `p` here not `head`
Your count idea id pretty interesting, though not as efficient it gets the job done.
thank you so much. now i can study my other classes. ^_^ THANK YOU!!!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question