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C++ Practice Exam 2022 Moed B

5782 - תשפב 2022 Moed B Semester Bet Answers:

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Part A:

Q1 B

  • cannot iterate over cat object, we need to define and overload a method to do so

Q2 A

  • First, it creates parent object (flower), then petal. Next it calls a method with value (not reference, so it needs the CC). Now, it utlizes the move constructor when returning the object… In the main method, the return value isn’t being used, so it destroys the instance as it went out of scope. Pf and A are destoryed. Then it prints “main”, and now the compiler knows that the program is about to end, so it calls the destructor on the rest of the items (in reverse order to creation).

Q3 D

  • A: This is how we declare pure virtual classes with = 0;

  • B: Abstract implies that there is one method that is pure virtual. And when there is a pure virtual function, you cannot make an instance of that class.

  • C: Yes, a single pure virtual method causes the whole class to be abstract.

  • D: False, any class that inherits from an abstract class and doesn’t implement those pure virtual functions is still abstract and legal.

  • E: Abstract base classes are useful for creating polymorphic programs. An abstract base class defines an interface without an implementation.

Q4 E

  • Static variables/members are on a per class level. Meaning, that B’s num is not the same as A’s num. In addition, because B uses a * (pointer), it doesn’t actually instantiate the class. Therefore, it is set to zero. Regarding A’s num, it instantiates 5 objects because of how we set it up in the array.

Q5 B

  • printVal() is a virtual function in addition to print(). Firstly, we call print() in the Father class which then calls printVal() also of the same object.

Q6 C

  • In the 1st program, we successfully write to the file. But we only write a single character and the rating. This is because ‘char * name;’ is used but not ‘char name[20];’, and as we know the pointer just links to the first element in the array.

  • Now, when we go to read from the file, it fails to access the Restaurant’s name field.

Q7 A

  • We are using the OR command with a 1, this essentially always passes a 1 into the Y variable in addition to whatever was held in X. Because it is also unsigned, it will always be >= 1. Now we shift that number 31 spots. This will result in an enourmous number which is 231 in decimal. And that translates to 2147483648 which is equal to 0x80000000 in Hex. As this either hits or exceeds the max limit of an INT, it will always be equal to that number.

View Code for Part A

Part B:


    // if at least 1 student has perfect gpa (avg of 100)
    if (any_of(students.begin(), students.end(), [](Student s)
               { return s.getAvg() >= 100; }))
        cout << count_if(students.begin(), students.end(), [](Student s)
                         { return s.getAvg() >= 90; })
             << endl;


  • A:

    Enter day: 0
    Enter season: 0
    exception in day
  • B:

    Cannot access member field

    הערך של months[0]->days לא יהיה מאותחל

      member "Month::days" (declared at line 13) is inaccessibleC/C++(265)


    #pragma once
    template <class T>
    class Calc {
            T a=0, b=0;
            T add();
            T sub();
            T mul();
            T div();
            Calculator(T a, T b);

    template <class T>
    T Calculator<T>::add() {
        return a+b;


  • A:

      void SearchTree::add(Node *current, int val, int level)
          if (current->value < val)
                  current->right = new Node(nullptr, nullptr,nullptr,level);
              else add(current->right, val, level + 1);
          else {
              if (!current->left)
                  current->left = new Node(nullptr, nullptr,nullptr,level);
              else add(current->left, val, level + 1);
  • B:

(connect method) may be incorrect and/or incomplete To implement the connect method in the binary search tree to connect all nodes at the same level to a linked list, we can use a breadth-first traversal approach using the STL queue. Each level of the tree will be traversed, and we will connect the nodes at the same level through their next pointers.

  void SearchTree::connect(){

     if (root == nullptr) {  //base case

        queue<node*> *nodeQueue = new queue<node*>();

        while (!nodeQueue.empty()) {
            int levelSize = nodeQueue.size();
            node* prev = nullptr;

            for (int i = 0; i < levelSize; i++) {
                node* current = nodeQueue.front();

                if (prev != nullptr) {
                    prev->next = current;

                prev = current;

                if (current->left) {

                if (current->right) {

View Code for Part B

Part C:

Q12 is in the link below!

View Code for Part C

Updated on August 7, 2022