function composition in C++ / C++11 - c++

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 :
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)).
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'
<< "5 == " << (f * g * h)(0) << '\n'
<< "100 == " << 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

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

Answer 5

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 6

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 7

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