How to search a linked list for data and return the data.

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09-21-2020
meagangramlin

How to search a linked list for data and return the data.

I know I am not supposed to post my homework on here but I asked my teacher for help but she won't answer me. For this program I am not allowed to use vectors or std::

I have a function called findStudent that takes a linked list structure as an argument. The function asks the user for a student's id number. I build a temporary structure object and search the linked list by calling the "retrieve" function. If the student is found them I am supposed to display the student's entire record. However, I can't get my function to find the student.

I use a structure named Student that stores all the student's info, and a linked list where each student's info is a node. In my structure I have a data member

Code:

int id

. In my linked list program, I call the function "retrieve" that is supposed to return the node's data I am searching for using the parameter.

This is my findStudent function:

Code:

void findStudent(const LinkedList<Student> &list) { int id; cout << "Please enter student ID number: "; cin >> id; Student temp{ id }; //temporary struct object if (list.retrieve(temp) == false) //can't find the student id i am searching for cout << "Student not found.\n"; else cout << temp; }

And this is the retrieve function I call above:

Code:

template <class TYPE> bool LinkedList<TYPE>::retrieve(TYPE &dataIn) const { bool success = false; Node<TYPE> *pTemp = front; while (pTemp != nullptr && pTemp->data < dataIn) { pTemp = pTemp->next; } if (pTemp != nullptr && pTemp->data == dataIn) { success = true; dataIn = pTemp->data; } return success; }

How can I get retrieve to find the student I am searching for?
09-21-2020
laserlight

Your linked list template's retrieve member function should have a const reference parameter instead:

Code:

template <class TYPE> bool LinkedList<TYPE>::retrieve(const TYPE &dataIn) const

The reason is that the act of finding if dataIn matches an element in the linked list conceptually does not change dataIn, so there's no reason for it to be a non-const reference parameter. What you're doing now with this:

Code:

dataIn = pTemp->data;

is conceptually useless because dataIn is already equal to pTemp->data. If what you want to do is actually "retrieve" the element as the name implies, then instead of returning a bool, you could have your function return an iterator (or a pointer) to the Student object, returning an end() iterator (or null pointer) if the Student object is not found.

Next, what you need to decide is how a generic "retrieve" member function should work to decide whether an element is or is not a match for the argument. What you're doing right now is assuming that objects of type TYPE can be compared by using operator< and operator==. This can work, but it means that you need to overload those operators for Student. Frankly, you don't need operator< unless the linked list is sorted, and if you're assuming that the linked list is sorted, there should be some indication that this is the case. So what I suggest is that you simplify your retrieve function to only compare by operator==, then overload operator== for Student to compare by id.
09-21-2020
meagangramlin

Ok so I updated my findStudent code to:

Code:

void findStudent(const LinkedList<Student> &list) { int id; cout << "Please enter student ID number: "; cin >> id; Student temp{ id }; list.retrieve(temp); cout << temp; }

And now my retrieve function looks like:

Code:

template <class TYPE> TYPE LinkedList<TYPE>::retrieve(TYPE &dataIn) const { Node<TYPE> *pTemp = front; while (pTemp != nullptr && pTemp->data < dataIn) { pTemp = pTemp->next; } if (pTemp != nullptr && pTemp->data == dataIn) { dataIn = pTemp->data; } return dataIn; }

I tried to make the parameter TYPE &dataIn a constant but my program would not work. But now I have another problem. Instead of printing out the student's id, name, gpa, and major it only prints out the id. This info is stored in the structure Student as:

Code:

struct Student { int id; char name[50]; float gpa; char major[6]; //overloaded operator functions }

How can I get it to display the student's entire record?
09-21-2020
laserlight

Quote:

Originally Posted by meagangramlin

I tried to make the parameter TYPE &dataIn a constant but my program would not work.

That's because you're using dataIn as an input/output parameter. I was trying to get you to redesign your program to use it as an input parameter, then provide the output by another way.

Quote:

Originally Posted by meagangramlin

But now I have another problem. Instead of printing out the student's id, name, gpa, and major it only prints out the id.

That's because now you're still using dataIn as an input/output parameter, except that you didn't actually assign to it in the function template.

Given what it looks like you want to do, i.e., assign to an output parameter, the way I would design this is to have retrieve be a function template itself that takes a predicate and has a TYPE& parameter as an output parameter, returning a bool to indicate found or not found:

Code:

template<typename T> template<typename Pred> bool LinkedList<T>::retrieve(T& result, Pred predicate) const { Node<T>* node = front; while (node) { if (predicate(node->data)) { result = node->data; return true; } node = node->next; } return false; }

You would then use it like this:

Code:

int id; // ... read into id ... Student result; if (students.retrieve(result, [id](const Student& s) { return id == s.id; })) { cout << result; } else { cout << "Student not found.\n"; }

Of course, this may be beyond the scope of your exercise. If so, what I suggest for you to just get this working though is to go back to your post #1, revert the code to that state, and then just overload operator== to compare Student ids. I also suggest renaming dataIn to something that reflects the fact that it is an input/output parameter, not an input parameter.