Question:
What is the output of the following C++ code?
# include <iostream.h>
class Base {
public:
void show() {cout << "Base"; }
};
class Derived : public Base {
public:
void show() {cout << "Derived"; }
};
int main() {
Base* b = new Derived();
b->show();
}
What is the output of the following C++ code?
# include <iostream.h>
class Base {
public:
void show() {cout << "Base"; }
};
class Derived : public Base {
public:
void show() {cout << "Derived"; }
};
int main() {
Base* b = new Derived();
b->show();
}
# include <iostream.h>
class Base {
public:
void show() {cout << "Base"; }
};
class Derived : public Base {
public:
void show() {cout << "Derived"; }
};
int main() {
Base* b = new Derived();
b->show();
}
Show Hint
Is show() declared virtual? Without virtual, a base pointer calls the base version.
Updated On: Jul 2, 2026
- Derived
- Base
- Compilation error
- Runtime error
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Look at show() in the Base class. It is a plain member function. It is not marked virtual.
Step 2: The call is made through a Base pointer:
Base* b = new Derived();
b->show();
Step 3: For a non virtual function, C++ uses static binding. The function to call is chosen from the static type of the pointer, which is Base, not from the actual object type Derived.
Step 4: So \( b\text{-}\!>show() \) resolves to Base::show at compile time and prints Base. The Derived version is ignored because there is no virtual dispatch.
Answer: option B.
Step 2: The call is made through a Base pointer:
Base* b = new Derived();
b->show();
Step 3: For a non virtual function, C++ uses static binding. The function to call is chosen from the static type of the pointer, which is Base, not from the actual object type Derived.
Step 4: So \( b\text{-}\!>show() \) resolves to Base::show at compile time and prints Base. The Derived version is ignored because there is no virtual dispatch.
Answer: option B.
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