Passing a functor as C++ template parameter - c++

Question

As an exercise for my personal enlightenment, I implement vector math with expression templates. I want to implement some operations that apply the same unary function to all elements to a vector expression. So far, I do this.
My base vector expression template is implemented like this
template <typename E>
class VectorExpr {
public:
int size() const { return static_cast<E const&>(*this).size(); }
float operator[](int i) const { return static_cast<E const&>(*this)[i]; }
operator E& () { return static_cast<E&>(*this); }
operator E const& () const { return static_cast<const E&>(*this); }
}; // class VectorExpr
Then, an object supposed to be a vector will look like this
class Vector2 : public VectorExpr<Vector2> {
public:
inline size_t size() const { return 2; }
template <typename E>
inline Vector2(VectorExpr<E> const& inExpr) {
E const& u = inExpr;
for(int i = 0; i < size(); ++i)
mTuple[i] = u[i];
}
private:
float mTuple[2];
};
Let's say I want to apply std::sin to all elements of an expression
template <typename E>
class VectorSin : public VectorExpr<VectorSin<E> > {
E const& mV;
public:
VectorSin(VectorExpr<E> const& inV) : mV(inV) {}
int size() const { return mV.size(); }
float operator [] (int i) const { return std::sin(mV[i]); }
};
Question => If I want to add more functions, I copy-paste what I do for the sin function, for every single function (like cos, sqrt, fabs, and so on). How I can avoid this kind of copy-pasting ? I tried things and figured out I'm still low in template-fu. No boost allowed ^^

Answer 1

template <typename F, typename E>
class VectorFunc : public VectorExpr<VectorFunc<F, E> > {
E const& mV;
public:
VectorSin(VectorExpr<E> const& inV) : mV(inV) {}
int size() const { return mV.size(); }
float operator [] (int i) const { return f(mV[i]); }
// this assumes the Functor f is default constructible, this is
// already not true for &std::sin. Adding the constructor that
// takes f, is left as an exercise ;)
F f;
};

Answer 2

In addition to the answer by pmr, The standard <cmath> functions aren't functors, so you couldn't use them directly to specify unique specialisations of your class - i.e. you wouldn't have a separate template instantiation for std::sin versus std::cos (which is what I gather you're aiming for? correct me if I've misunderstood you on that).
You could create a wrapper in order to map a function pointer to a distinct type, e.g.
#include <iostream>
template< void (*FuncPtr)() > struct Func2Type
{
void operator() () { FuncPtr(); }
};
void Hello() { std::cout << "Hello" << std::endl; }
void World() { std::cout << "world" << std::endl; }
int main()
{
Func2Type<Hello> test1;
Func2Type<World> test2;
test1();
test2();
}
That way you could use them as template arguments in the same way as a normal functor class