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C++ | Exception Handling | Question 4
  • Difficulty Level : Easy
  • Last Updated : 24 Nov, 2013

Output of following program

using namespace std;
class Base {};
class Derived: public Base {};
int main()
   Derived d;
   try {
       throw d;
   catch(Base b) {
        cout<<"Caught Base Exception";
   catch(Derived d) {
        cout<<"Caught Derived Exception";
   return 0;

(A) Caught Derived Exception
(B) Caught Base Exception
(C) Compiler Error

Answer: (B)

Explanation: If both base and derived classes are caught as exceptions then catch block of derived class must appear before the base class. If we put base class first then the derived class catch block will never be reached.

In Java, catching a base class exception before derived is not allowed by the compiler itself. In C++, compiler might give warning about it, but compiles the code.

Quiz of this Question

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