C ++ / C ++ 11中的函数组合

Question

I am currently coding some cryptographic algorithms in C++11 that require a lot of function compositions. There are 2 types of composition I have to deal with :

1. Compose a function on itself a variable number of times. Mathematically, for a certain function F, F^n(x) = (F^{n-1} o F)(x) = F^{n-1}(F(x)).

2. Compose different functions together. For example, for some functions f,g,h,i,j and k of the same type, I'll have f(g(h(i(j(k(x)))))).

In my case, I'm using the following definition of F :

const std::vector<uint8_t> F(const std::vector<uint8_t> &x);

I would like to compose this function on itself n times. I have implemented the composition in a simple recursive way which is working fine :

const std::vector<uint8_t> compose(const uint8_t n, const std::vector<uint8_t> &x)
{
    if(n > 1)
       return compose(n-1, F(x));

    return F(x);
}

For this case, is there a more efficient way an proper way to implement this composition using c++11 but without using BOOST ? It would be great to use this form if it is possible of course :

answer = compose<4>(F)(x); // Same as 'answer = F^4(x) = F(F(F(F(x))))'

For the second case, I would like to implement the composition of a variable number of functions. For a given set of functions F0, F1, ..., Fn having the same definition as F, is there an efficient and proper way to compose them where n is variable ? I think variadic template would be useful here, but I don't know how to use them in that case.

Thanks for your help.

Answer 1

Something along these lines, perhaps (untested):

template <typename F>
class Composer {
  int n_;
  F f_;
public:
  Composer(int n, F f) : n_(n), f_(f) {}

  template <typename T>
  T operator()(T x) const {
    int n = n_;
    while (n--) {
      x = f_(x);
    }
    return x;
  }
};

template <int N, typename F>
Composer<F> compose(F f) {
  return Composer<F>(N, f);
}

EDIT: And for the second case ( tested this time ):

#include <iostream>

template <typename F0, typename... F>
class Composer2 {
    F0 f0_;
    Composer2<F...> tail_;
public:
    Composer2(F0 f0, F... f) : f0_(f0), tail_(f...) {}

    template <typename T>
    T operator() (const T& x) const {
        return f0_(tail_(x));
    }
};

template <typename F>
class Composer2<F> {
    F f_;
public:
    Composer2(F f) : f_(f) {}

    template <typename T>
    T operator() (const T& x) const {
        return f_(x);
    }
};

template <typename... F>
Composer2<F...> compose2(F... f) {
    return Composer2<F...>(f...);
}

int f(int x) { return x + 1; }
int g(int x) { return x * 2; }
int h(int x) { return x - 1; }

int main() {
  std::cout << compose2(f, g, h)(42);
  return 0;
}

Answer 2

Thanks for the fun question, Gabriel of year 2013. Here is a solution. It works in c++14.

#include <functional>
#include <iostream>
using std::function;

// binary function composition for arbitrary types
template <class F, class G>
auto compose(F f, G g){
  return [f,g](auto x){return f(g(x));};
}

// for convienience
template <class F, class G>
auto operator*(F f, G g){
  return compose(f,g);
}

// composition for n arguments
template <class F, typename... Fs>
auto compose(F f, Fs&&... fs) {
  return f * compose(fs...);
}

// composition for n copies of f
template <int i, class F>
//must wrap chain in a struct to allow partial template specialization
struct multi { static F chain(F f){
    return f * multi<i-1,F>::chain(f);
}};
template <class F>
struct multi <2,F> { static F chain(F f){
    return f * f;
}};
template <int i, class F>
F compose(F f) {return multi<i,F>::chain(f); }

int main(int argc, char const *argv[]) {

  function<double(int)> f = [](auto i){return i + 3;};
  function<int(double)> g = [](auto i){return i * 2;};
  function<int(int)   > h = [](auto i){return i + 1;};

  std::cout
    << '\n' <<  "9 == " << compose(f,g,f)   (0)  \
    << '\n' <<  "9 == " << (f * g * h)      (0)  \
    << '\n' <<  "3 == " << compose<100>(h)  (0)  \
    << '\n';

  return 0;
}

You can define

Matrix compose(Matrix f, Matrix g);

or

Rotation compose(Rotation f, Rotation g);

to reuse this code for all sorts of things.

Answer 3

A very general example ( g++ -std=c++1y composition.cpp ):

// ---------------------------------------------------------
// "test" part
// ---------------------------------------------------------
int f(int a) { return 2*a; }
double g(int a) { return a+2.5; }
double h(double a) { return 2.5*a; }
double i(double a) { return 2.5-a; }

class Functor {
  double x;
public:
  Functor (double x_) :  x(x_) { }
  double operator() (double a) { return a*x; }
};

// ---------------------------------------------------------
// ---------------------------------------------------------
int main () {

  auto l1 = [] (double a) { return a/3; };
  auto l2 = [] (double a) { return 3.5+a; };

  Functor fu {4.5};

  auto compos1 = compose (f, g, l1, g, h, h, l1, l2); 

  auto compos2 = compose (compos1, l1, l2, fu);

  auto x = compos2 (3);

  cout << x << endl;
  cout << compos2(3) << endl;

  cout << fu(l2(l1(l2(l1(h(h(g(l1(g(f(3))))))))))) << endl;

} // ()

Library part:

// ---------------------------------------------------------
// "library" part
// ---------------------------------------------------------
template<typename F1, typename F2>
class Composite{
private:
  F1  f1;
  F2  f2;

public:
  Composite(F1  f1,  F2  f2) : f1(f1), f2(f2) { }

  template<typename IN>
  decltype(auto) operator() (IN i)   
  { 
    return f2 ( f1(i) ); 
  }
};

// ---------------------------------------------------------
// ---------------------------------------------------------
template<typename F1, typename F2>
decltype(auto) compose (F1 f, F2 g) {
  return Composite<F1, F2> {f,g};
}

// ---------------------------------------------------------
// ---------------------------------------------------------
template<typename F1, typename... Fs>
decltype(auto) compose (F1  f,  Fs  ... args) 
{
  return compose (f, compose(args...));
}

The whole program:

//  g++ -std=c++1y composition.cpp 

#include <iostream>

using namespace std;

// ---------------------------------------------------------
// "library" part
// ---------------------------------------------------------
template<typename F1, typename F2>
class Composite{
private:
  F1  f1;
  F2  f2;

public:
  Composite(F1  f1,  F2  f2) : f1(f1), f2(f2) { }

  template<typename IN>
  decltype(auto) operator() (IN i)   
  { 
    return f2 ( f1(i) ); 
  }
};

// ---------------------------------------------------------
// ---------------------------------------------------------
template<typename F1, typename F2>
decltype(auto) compose (F1 f, F2 g) {
  return Composite<F1, F2> {f,g};
}

// ---------------------------------------------------------
// ---------------------------------------------------------
template<typename F1, typename... Fs>
decltype(auto) compose (F1  f,  Fs  ... args) 
{
  return compose (f, compose(args...));
}

// ---------------------------------------------------------
// "test" part
// ---------------------------------------------------------
int f(int a) { return 2*a; }
double g(int a) { return a+2.5; }
double h(double a) { return 2.5*a; }
double i(double a) { return 2.5-a; }

class Functor {
  double x;
public:
  Functor (double x_) :  x(x_) { }
  double operator() (double a) { return a*x; }
};

// ---------------------------------------------------------
// ---------------------------------------------------------
int main () {

  auto l1 = [] (double a) { return a/3; };
  auto l2 = [] (double a) { return 3.5+a; };

  Functor fu {4.5};

  auto compos1 = compose (f, g, l1, g, h, h, l1, l2); 

  auto compos2 = compose (compos1, l1, l2, fu);

  auto x = compos2 (3);

  cout << x << endl;
  cout << compos2(3) << endl;

  cout << fu(l2(l1(l2(l1(h(h(g(l1(g(f(3))))))))))) << endl;

} // ()

Answer 4

A quick implementation of function iteration with argument forwarding. The helper type is unfortunately necessary because function templates can't be partially specialised.

#include <functional>
#include <iostream>
using namespace std;

template<int n, typename A>
struct iterate_helper {
  function<A(A)> f;
  iterate_helper(function<A(A)> f) : f(f) {}
  A operator()(A&& x) {
    return f(iterate_helper<n - 1, A>(f)(forward<A>(x)));
  };
};

template<typename A>
struct iterate_helper<1, A> {
  function<A(A)> f;
  iterate_helper(function<A(A)> f) : f(f) {}
  A operator()(A&& x) {
    return f(forward<A>(x));
  };
};

template<int n, typename A>
function<A(A)> iterate(function<A(A)> f) {
  return iterate_helper<n, A>(f);
}

int succ(int x) {
  return x + 1;
}

int main() {
  auto add5 = iterate<5>(function<int(int)>(succ));
  cout << add5(10) << '\n';
}

Answer 5

You haven't shown the body of F , but if you can modify it so that it mutates the input to form the output then change the signature to:

void F(std::vector<uint8_t>& x);

Thereafter you can implement Fn as:

void Fn(std::vector<uint8_t>& x, size_t n)
{
    for (size_t i = 0; i < n; i++)
        F(x);
}

The compiler will unroll the loop for you if it is more efficient, but even if it doesn't an increment/compare of a local variable will be orders of magnitude faster than calling F.

You can then explcitly copy-construct new vectors when you actually want to make a copy:

vector<uint8_t> v1 = ...;
vector<uint8_t> v2 = v1; // explicitly take copy
Fn(v2,10);

Answer 6

What about (untested):

template < typename Func, typename T >
T  compose_impl( Func &&, T &&x, std::integral_constant<std::size_t, 0> )
{ return std::forward<T>(x); }

template < typename Func, typename T, std::size_t N >
T compose_impl( Func &&f, T &&x, std::integral_constant<std::size_t, N> )
{
    return compose_impl( std::forward<Func>(f),
     std::forward<Func>(f)(std::forward<T>( x )),
     std::integral_constant<std::size_t, N-1>{} );
}

template < std::size_t Repeat = 1, typename Func, typename T >
T  compose( Func &&f, T &&x )
{
    return compose_impl( std::forward<Func>(f), std::forward<T>(x),
     std::integral_constant<std::size_t, Repeat>{} );
}

We can use variadic function templates for multiple functions (untested):

template < typename Func, typename T >
constexpr  // C++14 only, due to std::forward not being constexpr in C++11
auto chain_compose( Func &&f, T &&x )
 noexcept( noexcept(std::forward<Func>( f )( std::forward<T>(x) )) )
 -> decltype( std::forward<Func>(f)(std::forward<T>( x )) )
{ return std::forward<Func>(f)(std::forward<T>( x )); }

template < typename Func1, typename Func2, typename Func3, typename ...RestAndT >
constexpr  // C++14 only, due to std::forward
auto  chain_compose( Func1 &&f, Func2 &&g, Func3 &&h, RestAndT &&...i_and_x )
 noexcept( CanAutoWorkHereOtherwiseDoItYourself )
 -> decltype( auto )  // C++14 only
{
    return chain_compose( std::forward<Func1>(f),
     chain_compose(std::forward<Func2>( g ), std::forward<Func3>( h ),
     std::forward<RestAndT>( i_and_x )...) );
}

The upcoming decltype(auto) construct automatically computes the return type from an inlined function. I don't know if there's a similar automatic computation for noexcept

Answer 7

Here is a simple c++14 solution (it may probably be re-written to c++11):

#include <iostream>

// base condition
template <typename F>
auto compose(F&& f)
{
    return [a = std::move(f)](auto&&... args){
        return a(std::move(args)...);
    };
}

// recursive composition
// from compose(a, b, c...) to compose(ab, c...)
template <typename F1, typename F2, typename... Fs>
auto compose(F1&& f1, F2&& f2, Fs&&... fs)
{
    return compose(
        [first = std::move(f1), second = std::move(f2)]
        (auto&&... args){
            return second(first(std::move(args)...));
        },
        std::move(fs)...
    );
}

Possible usage:

int main()
{
const auto f = compose(
  [](const auto n){return n * n;},
  [](const auto n){return n + 2;},
  [](const auto n){return n + 2;}
);
std::cout << f(10) << std::endl; // outputs 104
}

Here is a link to the repo with a few more examples: https://github.com/nestoroprysk/FunctionComposition