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    Question

    Consider the following C++ code:     class Base

    {     public:         void show() { std::cout
    A Base::show Correct Answer Incorrect Answer
    B Derived::show Correct Answer Incorrect Answer
    C Compilation Error Correct Answer Incorrect Answer
    D Runtime Error Correct Answer Incorrect Answer
    E Nothing, the program will not execute. Correct Answer Incorrect Answer

    Solution

    In C++, if a base class method is not declared virtual, method calls through a base class pointer (or reference) are resolved at compile time based on the type of the pointer(Base), not the actual object it points to (Derived). To achieve Derived::show output, Base::show() would need to be declared virtual.

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