How can I make the method of child be called: virtual keyword not working?

Question

The following is my code,

#include
#include

using namespace std;

class TestClass
{
  public:
    virtual void test(string st1         


        

Answer 1

void pass(TestClass t)
{
    t.test("abc","def");
}

When you do this, the object you are passing gets sliced into a TestClass and its identity is lost, so it now behaves like a TestClass and calls the method accordingly.

To Fix this you want to pass t by reference as @Nick suggested, or (not recommended) by pointer. It will now retain its identity and call the appropriate function, as long as test is marked virtual

Edit: fixed spliced -> sliced .. too much bioshock..

Answer 2

As noted above, this is a result of "slicing". The specific details of what occurs is as follows:

When arguments are passed by value, a copy of the argument is passed to the function. When this happens, a new object is constructed by calling the copy-constructor.

For your example, the copy-constructor has a signature as follows:

TestClass::TestClass(const TestClass&);

so, what really happens is something like the following (again, for your example):

ExtendedTest et();
pass(et);
{ // entering scope of pass function ...
  TestClass t = TestClass(t_orig); // inserted by the compiler
  // evaluation of pass function ...
  // ...
} // leaving scope of pass function, t is destroyed.

Obviously, since the variable t is instantiated as a TestClass, any member function calls will be from TestClass (and not ExtendedTest).

As a final note, you should always declare virtual destructors when using inheritance. This will avoid slicing when objects are destroyed.

Answer 3

You need to change the parameter for a reference (or a pointer)

void pass(TestClass &t)

This way, the original object will be used.