Operator composition in c++ - 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;
}

Answer 1

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;
}

Answer 2

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.

Answer 3

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

Answer 4

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.