Question

The following is my code,

#include<iostream>
#include<string>

using namespace std;

class TestClass
{
  public:
    virtual void test(string st1, string st2);

};

class ExtendedTest: public TestClass
{
  public:
    virtual void test(string st1, string st2);
};

void TestClass::test(string st1, string st2="st2")
{
     cout << st1 << endl;
     cout << st2 << endl;
}

void ExtendedTest::test(string st1, string st2="st2")
{
     cout << "Extended: " << st1 << endl;
     cout << "Extended: " << st2 << endl;
}

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


int main()
{
   ExtendedTest et;
   pass(et);
   return 0;
}

When I run the code, the method(‘test’) of base class is called.
But I expect the method of child is called because I specified methods as virtual function.

Then how can I make method of child class be called ? Thank you.

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..