How to make return type of a function to be same as another function?

there is a question in our study book about object functions. There is a code in c++ and the question wants us to fill the blanks. The code is as below

template <typename Arg, typename Ret> 
class FuncObj {
public:
    typedef Arg argType;
    typedef Ret retType;
    virtual Ret operator()(Arg) = 0;
};

class DivideBy : public FuncObj<int, double> {
protected:
    int divisor;
public:
    DivideBy(int d) {
        this->divisor = d;
    }
    double operator()(int x) {
        return x/((double)divisor);
    }
};

class Truncate : public FuncObj<double, int> {
public:
    int operator()(double x) {
        return (int) x;
    }
};

template < typename Ftype , typename Gtype >
class Compose : public FuncObj <typename Gtype :: argType, typename Ftype :: retType > {
protected:
    Ftype *f; Gtype *g;
public:
    Compose(Ftype f,Gtype g) {
    --------- =f;
    --------- =g; 
    }

    ---------- operator()(____________x) {
        return (_________)((________)(__________)); }
};

The desirable result is

void main() {
    DivideBy *d = new DivideBy(2);
    Truncate *t = new Truncate();
    Compose<DivideBy, Truncate> *c1 = new Compose<DivideBy,Truncate>(d,t);
    Compose<Truncate, DivideBy> *c2 = new Compose<Truncate, DivideBy>(t,d);
    cout << (*c1)(100.7) << endl; // Prints 50.0 
    cout << (*c2)(11) << endl; // Prints 5
}

I really don't know how to complete this code, so what feature or concept of c++ should we use to make this work? If there is a link for further study about this topic please write it down. thanks.