Operator composition in c++

Question

I'm wondering if there is an elegant solution for composing mathematical operators in C++. By operator, I mean something like the following:

template<class H>
class ApplyOp {
    H h;
public:
    ApplyOp(){}
    ApplyOp(H h_i) : h(h_i) {}

    template<class argtype>
    double operator()(argtype f,double x){
        return h(x)*f(x);
    }
};

The above class makes use of a "helper function" h(x). For example,

struct Helper{
    Helper(){}
    double operator()(double x){return x*x;}
};

struct F{
    F(){}
    double operator()(double x){return exp(x);}
};

int main()
{
    Helper h;
    F f;
    ApplyOp<Helper> A(h);

    std::cout<<"A(f,2.0) = "<<A(f,2.0)<<std::endl; //Returns 2^2*exp(2) = 29.5562...

    return 0;
}

Now, I would like to compose the operator twice or more times, i.e. compute A^2(f,2.0). In the above example, this would return h(x)*h(x)*f(x). Note that this is not function composition, i.e. I do not want to compute A(A(f,2.0),2.0). Rather, think in terms of computing powers of a matrix: if h(x) = M (a matrix), I want M*M*...*M*x.

I was able to use std::bind() to achieve my desired result for A^2 (but not higher powers!) as follows:

auto g = std::bind(&ApplyOp<Helper>::operator()<F>,&A,f,std::placeholders::_1);

With the resulting g, I can apply A^2(f,2.0) by simply calling A(g,2.0). With the above examples, this would return h(x)*h(x)*f(x) = x*x*x*x*exp(x)

How would I generalize this to iteratively applying the operator A N times? I really liked the answer posted here, but it doesn't quite work here. I tried doing nested std:binds but quickly got into deep compiler errors.

Any ideas?

Complete working example:

#include<iostream>
#include<math.h>
#include<functional> //For std::bind

template<class H>
class ApplyOp {
    H h;
public:
    ApplyOp(){}
    ApplyOp(H h_i) : h(h_i) {}

    template<class argtype>
    double operator()(argtype f,double x){
        return h(x)*f(x);
    }
};

struct Helper{
    Helper(){}
    double operator()(double x){return x*x;}
};

struct F{
    F(){}
    double operator()(double x){return exp(x);}
};

int main()
{
    Helper h;
    F f;
    ApplyOp<Helper> A(h);

    std::cout<<"A(f,2.0) = "<<A(f,2.0)<<std::endl; //Returns 2^2*exp(2) = 29.5562...

    auto g = std::bind(&ApplyOp<Helper>::operator()<F>,&A,f,std::placeholders::_1);

    std::cout<<"A^2(f,2.0) = "<<A(g,2.0) <<std::endl; //Returns 2^4*exp(2) = 118.225... 

    return 0;
}

to have `h(x)*h(x)*f(x)`, you have to change to have `h1(x) = h(x)*h(x)` and use `ApplyOp A(h1);`. — Jarod42, Oct 05 '18 at 16:50
@Jarod42 that's somewhat missing the point. I want a more general solution that doesn't depend on the particular form of the operator. — icurays1, Oct 05 '18 at 17:18
@user9335240 I'm not sure I see how specialization helps - can you explain a little more? — icurays1, Oct 05 '18 at 17:19
@Jarod42 You can make a template function that has a template parameter of an int, that recursively resolves to the same function but with the template parameter subtracted by one, and make a specialization when this parameter == 2 that does the binding — user9335240, Oct 05 '18 at 17:21

lakshayg · Answer 1 · 2018-10-05T20:52:25.790

From what I could understand from you question, you are essentially trying to define

A^1(h, f, x) = h(x) * f(x)
A^n(h, f, x) = h(x) * A^(n-1)(h, f, x)

If you are open to using C++17, here is something you can build upon:

#include <iostream>
#include <math.h>

template <int N>
struct apply_n_helper {
  template <typename H, typename F>
  auto operator()(H h, F f, double x) const {
    if constexpr(N == 0) {
      return f(x);
    } else {
      return h(x) * apply_n_helper<N - 1>()(h, f, x);
    }
  }
};

template <int N>
constexpr auto apply_n = apply_n_helper<N>();

int main() {
  auto sqr = [](double x) { return x * x; };
  auto exp_ = [](double x) { return exp(x); };

  std::cout << apply_n<100>(sqr, exp_, 2.0) << '\n';
  std::cout << apply_n<200>(sqr, exp_, 2.0) << '\n';
  return 0;
}

If C++17 is not an option, you can easily rewrite this to use template specializations instead of constexpr-if. I will leave this as an exercise. Here is a link to compiler explorer with this code: https://godbolt.org/z/5ZMw-W

EDIT Looking back at this question, I see that you are essentially trying to compute (h(x))^n * f(x) in a manner so that you don't have to actually do any looping at runtime and the generated code is equivalent to something like:

auto y = h(x);
auto result = y * y * ... * y * f(x)
              \_____________/
                  n times
return result;

Another way of achieving this would be to have something as follows

#include <cmath>
#include <iostream>

template <size_t N, typename T>
T pow(const T& x) {
    if constexpr(N == 0) {
        return 1;
    } else if (N == 1) {
        return x;
    } else {
        return pow<N/2>(x) * pow<N - N/2>(x);
    }
}

template <int N>
struct apply_n_helper {
    template <typename H, typename F>
    auto operator()(H h, F f, double x) const {
        auto tmp = pow<N>(h(x));
        return tmp * f(x);
    }
};

template <int N>
constexpr auto apply_n = apply_n_helper<N>();

int main()
{
    auto sqr = [](double x) { return x * x; };
    auto exp_ = [](double x) { return exp(x); };

    std::cout << apply_n<100>(sqr, exp_, 2.0) << '\n';
    std::cout << apply_n<200>(sqr, exp_, 2.0) << '\n';
    return 0;
}

Here, the usage of pow function is saving us from evaluating h(x) several times.

score 1 · Accepted Answer · answered Oct 05 '18 at 19:24

Try this, using template specialization

#include<iostream>
#include<math.h>
#include<functional> //For std::bind

template<class H>
class ApplyOp {
    H h;
public:
    ApplyOp(){}
    ApplyOp(H h_i) : h(h_i) {}

    template<class argtype>
    double operator()(argtype f,double x){
        return h(x)*f(x);
    }
};

struct Helper{
    Helper(){}
    double operator()(double x){return x*x;}
};

struct F{
    F(){}
    double operator()(double x){return exp(x);}
};

// C++ doesn't permit recursive "partial specialization" in function
// So, make it a struct instead
template<typename T, typename U, typename W, int i>
struct Binder {
    auto binder(U b, W c) {
        // Recursively call it with subtracting i by one
        return [&](T x){ return b(Binder<T, U, W, i-1>().binder(b, c), x); };
    }
};

// Specialize this "struct", when i = 2
template<typename T, typename U, typename W>
struct Binder<T, U, W, 2> {
    auto binder(U b, W c) {
        return [&](T x){ return b(c, x); };
    }
};

// Helper function to call this struct (this is our goal, function template not
// struct)
template<int i, typename T, typename U, typename W>
auto binder(U b, W d) {
    return Binder<T, U, W, i>().binder(b, d);
}

int main()
{
    Helper h;
    F f;
    ApplyOp<Helper> A(h);

    std::cout<<"A(f,2.0) = "<<A(f,2.0)<<std::endl; //Returns 2^2*exp(2) = 29.5562...

    // We don't need to give all the template parameters, C++ will infer the rest
    auto g = binder<2, double>(A, f);

    std::cout<<"A^2(f,2.0) = "<<A(g,2.0) <<std::endl; //Returns 2^4*exp(2) = 118.225... 

    auto g1 = binder<3, double>(A, f);

    std::cout<<"A^3(f,2.0) = "<<A(g1,2.0) <<std::endl; //Returns 2^6*exp(2) = 472.2

    auto g2 = binder<4, double>(A, f);

    std::cout<<"A^4(f,2.0) = "<<A(g2,2.0) <<std::endl; //Returns 2^8*exp(2) = 1891.598... 

    return 0;
}

This is _extremely_ close to what I need. I tried to implement it in my project and it's not quite working - for some reason, the lambda returned by `binder` isn't capturing the correct behavior. If I do `auto g = [&] (double x){return A(f,x);};` then evaluate `A(g,x)`, I get the correct result, but not if I use `auto g = binder<2,double>(A,f)`. It works brilliantly for the example I gave (your code works), but when I make the change to my `ApplyOp` it breaks. I may post a separate question as it's difficult to explain in a comment. Thanks! — icurays1, Oct 05 '18 at 21:32
OK, this turned out to be _exactly_ what I needed. I needed to capture by value when returning the lambda expression (i.e. use `[=]`), and make sure all my `operator()` were `const`. Great solution, thanks again! — icurays1, Oct 05 '18 at 22:27

Yakk - Adam Nevraumont · Answer 3 · 2018-10-05T19:03:06.047

So you want to be able to multiply functions. Well, that sounds good. Why not + and - and / while we are in there?

template<class F>
struct alg_fun;

template<class F>
alg_fun<F> make_alg_fun( F f );

template<class F>
struct alg_fun:F {
  alg_fun(F f):F(std::move(f)){}
  alg_fun(alg_fun const&)=default;
  alg_fun(alg_fun &&)=default;
  alg_fun& operator=(alg_fun const&)=default;
  alg_fun& operator=(alg_fun &&)=default;

  template<class G, class Op>
  friend auto bin_op( alg_fun<F> f, alg_fun<G> g, Op op ) {
    return make_alg_fun(
      [f=std::move(f), g=std::move(g), op=std::move(op)](auto&&...args){
        return op( f(decltype(args)(args)...), g(decltype(args)(args)...) );
      }
    );
  }

  template<class G>
  friend auto operator+( alg_fun<F> f, alg_fun<G> g ) {
    return bin_op( std::move(f), std::move(g), std::plus<>{} );
  }
  template<class G>
  friend auto operator-( alg_fun<F> f, alg_fun<G> g ) {
    return bin_op( std::move(f), std::move(g), std::minus<>{} );
  }
  template<class G>
  friend auto operator*( alg_fun<F> f, alg_fun<G> g ) {
    return bin_op( std::move(f), std::move(g),
      std::multiplies<>{} );
  }
  template<class G>
  friend auto operator/( alg_fun<F> f, alg_fun<G> g ) {
    return bin_op( std::move(f), std::move(g),
      std::divides<>{} );
  }

  template<class Rhs,
    std::enable_if_t< std::is_convertible<alg_fun<Rhs>, F>{}, bool> = true
  >
  alg_fun( alg_fun<Rhs> rhs ):
    F(std::move(rhs))
  {}

  // often doesn't compile:
  template<class G>
  alg_fun& operator-=( alg_fun<G> rhs )& {
    *this = std::move(*this)-std::move(rhs);
    return *this;
  }
  template<class G>
  alg_fun& operator+=( alg_fun<G> rhs )& {
    *this = std::move(*this)+std::move(rhs);
    return *this;
  }
  template<class G>
  alg_fun& operator*=( alg_fun<G> rhs )& {
    *this = std::move(*this)*std::move(rhs);
    return *this;
  }
  template<class G>
  alg_fun& operator/=( alg_fun<G> rhs )& {
    *this = std::move(*this)/std::move(rhs);
    return *this;
  }
};
template<class F>
alg_fun<F> make_alg_fun( F f ) { return {std::move(f)}; }

auto identity = make_alg_fun([](auto&& x){ return decltype(x)(x); });
template<class X>
auto always_return( X&& x ) {
  return make_alg_fun([x=std::forward<X>(x)](auto&&... /* ignored */) {
    return x;
  });
}

I think that about does it.

auto square = identity*identity;

we can also have type-erased alg funs.

template<class Out, class...In>
using alg_map = alg_fun< std::function<Out(In...)> >;

these are the ones that support things like *=. alg_fun in general don't type erase enough to.

template<class Out, class... In>
alg_map<Out, In...> pow( alg_map<Out, In...> f, std::size_t n ) {
  if (n==0) return always_return(Out(1));
  auto r = f;
  for (std::size_t i = 1; i < n; ++i) {
    r *= f;
  }
  return r;
}

could be done more efficiently.

Test code:

auto add_3 = make_alg_fun( [](auto&& x){ return x+3; } );
std::cout << (square * add_3)(3)  << "\n";; // Prints 54, aka 3*3 * (3+3)

alg_map<int, int> f = identity;
std::cout << pow(f, 10)(2) << "\n"; // prints 1024

Live example.

Here is a more efficient pow that works without type erasure:

inline auto raise(std::size_t n) {
  return make_alg_fun([n](auto&&x)
    -> std::decay_t<decltype(x)>
  {
    std::decay_t<decltype(x)> r = 1;
    auto tmp = decltype(x)(x);

    std::size_t bit = 0;
    auto mask = n;
    while(mask) {
      if ( mask & (1<<bit))
        r *= tmp;
      mask = mask & ~(1<<bit);
      tmp *= tmp;
      ++bit;
    }
    return r;
  });
}
template<class F>
auto pow( alg_fun<F> f, std::size_t n ) {
  return compose( raise(n), std::move(f) );
}

Live example. It uses a new function compose in alg_fun:

  template<class G>
  friend auto compose( alg_fun lhs, alg_fun<G> rhs ) {
    return make_alg_fun( [lhs=std::move(lhs), rhs=std::move(rhs)](auto&&...args){
        return lhs(rhs(decltype(args)(args)...));
    });
  }

which does compose(f,g)(x) := f(g(x))

Your code now literally becomes

alg_fun<Helper> h;
alg_fun<F> f;

auto result = pow( h, 10 )*f;

which is h(x)*h(x)*h(x)*h(x)*h(x)*h(x)*h(x)*h(x)*h(x)*h(x)*f(x). Except (with the efficient version) I only call h once and just raise the result to the power 10.

score 0 · Answer 4 · answered Oct 05 '18 at 18:59

I mean you can use a other class:

template <typename T, std::size_t N>
struct Pow
{
    Pow(T t) : t(t) {}

    double operator()(double x) const
    {
        double res = 1.;

        for (int i = 0; i != N; ++i) {
            res *= t(x);
        }
        return res;
    }

    T t;  
};

And use

ApplyOp<Pow<Helper, 2>> B(h); instead of ApplyOp<Helper> A(h);

Demo

Operator composition in c++

4 Answers4