Selecting a child class problem

Hi,

I have an abstract class called Integrator and two child classes, EmbedRK and Rungekutta. In my main.cc what I want to do is define a variable rk as the type that I want to use (either EmbedRK and Rungekutta) depending on a config file. I have the following:

Code:

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const string integrator_type = config.read<string>( "Integrator" );
cout << integrator_type << endl;
if ( integrator_type == "embedrk" )       
{               
    cout << "Integrator: Adaptive step RungeKutta" << endl;               
    const double startstep = config.read<double>( "StartStep" );               
    cout << startstep << endl;               
    EmbedRK rk(equation, startstep, tolerance);       
}       
else if ( integrator_type == "rungekutta" )       
{               
      cout << "Integrator: RungeKutta" << endl;               
      Rungekutta rk(equation, tolerance);       
}             
else       
{               
      cout << "Integrator not recognised .. exiting" << endl;               
      exit(1);       
}

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The problem is that when I compile it, it complains when I use rk that 'rk' is not defined. I use it like this:

Code:

---

Model model_2(rk,equation);

---

where in this case the function is declared as follows:

Code:

---

Model(const Integrator &, Equations *);

---

Any help to sort this out would be great.

Daniel Brewer

One solution is RTTI. Pass the base case in the first parameter of model_2().

Kuphryn

Your problem is that you are declaring "rk" inside a block statement. That means it immediately goes out of scope at the end of that block. Your only real solution is to use a pointer:

Code:

---

const string integrator_type = config.read<string>( "Integrator" );
cout << integrator_type << endl;

Integrator* rk;

if ( integrator_type == "embedrk" )
{               
    cout << "Integrator: Adaptive step RungeKutta" << endl;
    const double startstep = config.read<double>( "StartStep" );
    cout << startstep << endl;
    rk = new EmbedRK(equation, startstep, tolerance);
}       
else if ( integrator_type == "rungekutta" )
{               
      cout << "Integrator: RungeKutta" << endl;
      rk = new Rungekutta(equation, tolerance);
}
else
{
      cout << "Integrator not recognised .. exiting" << endl;
      exit(1);
}

Model model2(*rk, equation);

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