Tech Mahindra Interview Questions and Answers

Diamond Problem occurs in multiple inheritance when a class inherits from two classes that have a common base class.

• Diamond Problem arises when a class inherits from two classes that have a common base class.

• Ambiguity arises in accessing the common base class members.

• To solve Diamond Problem, virtual inheritance is used to ensure only one copy of the common base class is inherited.

Constructors are special member functions in a class used to initialize objects. Copy constructors create a new object as a copy of an existing object.

• Types of constructors: Default constructor, Parameterized constructor, Copy constructor, and Destructor

• Types of copy constructors: Shallow copy constructor and Deep copy constructor

• Shallow copy constructor copies the values of the members of one object to another object. Example: MyClass(const MyClass &obj) { x = obj.x; y = obj...read more

Virtual functions in C++ allow for dynamic polymorphism by enabling a function to be overridden in a derived class.

• Virtual functions are used in C++ to achieve runtime polymorphism, where the function to be called is determined at runtime based on the object's type.

• They are needed to enable a base class pointer to call a function that is overridden in a derived class.

• Virtual functions help in achieving the Open/Closed principle of object-oriented programming, allowing for ext...read more

Smart pointers are objects that manage the memory of dynamically allocated objects in C++ to prevent memory leaks.

• Smart pointers automatically delete the object they point to when they are no longer needed.

• Examples include unique_ptr, shared_ptr, and weak_ptr in C++.

• They help in preventing memory leaks and dangling pointers.

• Smart pointers provide better memory management compared to raw pointers.

To find the middle of a linked list, use two pointers - one moving at double the speed of the other.

• Initialize two pointers, slow and fast, both pointing to the head of the linked list.

• Move the slow pointer by one node and the fast pointer by two nodes in each iteration.

• When the fast pointer reaches the end of the list, the slow pointer will be at the middle node.