How do one throw polymorphically?, C/C++ Programming

How do one throw polymorphically?              

A: Sometimes people write code such as:

class MyExceptionBase { };

class MyExceptionDerived : public MyExceptionBase { };

void f(MyExceptionBase& e)

{

// ... throw e;

}

void g()

{

MyExceptionDerived e;

try {

f(e);

}

catch (MyExceptionDerived& e) {

...code to handle MyExceptionDerived...

}

catch (...) {

...code to handle other exceptions...

}

}

If you attempt this, you might be astounded at run-time while your catch (...) clause is entered, & not your catch (MyExceptionDerived&) clause. It happens because you didn't throw polymorphically. The statement throw e , in function f(); throws an object  along with the same type as the static type of the expression e. In other terms, it throws an example of MyExceptionBase. The throw statement acts as-if the thrown object is copied, as opposed to developing a "virtual copy".

Luckily it's comparatively easy to correct:

class MyExceptionBase {

public:

virtual void raise();

};

void MyExceptionBase::raise()

{ throw *this; }

class MyExceptionDerived : public MyExceptionBase {

public:

virtual void raise();

};

 

void MyExceptionDerived::raise()

{ throw *this; }

void f(MyExceptionBase& e)

{

// ... e.raise();

}

void g()

{

MyExceptionDerived e;

try {

f(e);

}

catch (MyExceptionDerived& e) {

...code to handle MyExceptionDerived...

}

catch (...) {

...code to handle other exceptions...

}

}

Note down that the throw statement has been moved in a virtual function. The statement e.raise() will show polymorphic behavior, as raise() is declared virtual & e was passed through reference. As before, the thrown object will be of static type of the argument in the throw statement, however in MyExceptionDerived::raise(), that static type is MyExceptionDerived not MyExceptionBase.

 

Posted Date: 3/19/2013 8:52:07 AM | Location : United States







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