C ++ 11中具有可变参数模板的功能组合

Question

I'm a mathematician used to doing "old style" C++ programming for a long time now. I feel that some new syntactic constructions offerred by C++11 could help me achieve some better code regarding my professionnal projects. Yet as I'm no CS professionnal I must confess that I lack the knowledge to understand some examples I encounter in my self-learning process, altough I've been pretty lucky/succesful so far.

My impression is that variadic templates can be used to implement type-safe functions composition, as in this question . My concern is slightly more general since I'd like to compose functions with heterogeneous (but compatible) argument/return types. I've googled a lot and found another reference , but it seems utter "black magic" to me ;) and I won't pretend I can adapt the code in my context, although I feel I should find what I need in there.

I think the (most incomplete) code below is relatively self-explanatory as to what I'd like to achieve. In particular I believe the proper implementation will throw a compile-time error when one's trying to compose incompatible functions (Arrow here), and will need a piece of recursive template code.

template <typename Source , typename Target> class Arrow
{
  Target eval (const Source &);
};

template <typename ...Arrows> class Compositor
{
  template <typename ...Arrows>
  Compositor (Arrows... arrows)
  {
     // do/call what needs be here
  };

  auto arrow(); // gives a function performing the functionnal composition of arrows

};

// define some classes A, B and C

int main(int argc, char **argv)
{
  Arrow < A , B >  arrow1;
  Arrow < B , C >  arrow2;

  Compositor< Arrow < A , B > , Arrow < B , C > > compositor(arrow1 , arrow2);

  Arrow < A , C >  expected_result = compositor.arrow();
}

Ideally I'd like
Compositor
to directly subclass
Arrow < source_of_first_arrow , target_of_last_arrow>
and the method
arrow()
be replaced by the corresponding
eval()

but I felt the above code was more explanatory.

Any help will be greatly appreciated, even if it consists in a rough rebuke with a pointer to an existing (relatively basic) piece of example which will surely have escaped my search. Thanks!

Answer 1

If I got it correctly, you need no fancy template magic to do this composition. Here is the almost self-explanatory code:

#include <functional>
#include <string>
#include <iostream>

// it is just an std::function taking A and returning B
template <typename A, typename B>
using Arrow = std::function<B(const A&)>;

// composition operator: just create a new composed arrow
template <typename A, typename B, typename C>
Arrow<A, C> operator*(const Arrow<A, B>& arr1, const Arrow<B, C>& arr2)
{
    // arr1 and arr2 are copied into the lambda, so we won't lose track of them
    return [arr1, arr2](const A& a) { return arr2(arr1(a)); };
}

int main()
{
    Arrow<int, float> plusHalf([](int i){return i + 0.5;});
    Arrow<float, std::string> toString([](float f){return std::to_string(f);});

    auto composed = plusHalf * toString; // composed is Arrow<int, std::string>
    std::cout << composed(6) << std::endl; // 6.5

    //auto badComposed = toString * plusHalf; // compile time error
}

I mostly played with lambda functions here.

Using a single function call instead of a operator chain proved to be a more tricky problem. This time you got some templates:

#include <functional>
#include <string>
#include <iostream>

// it is just an std::function taking A and returning B
template <typename A, typename B>
using Arrow = std::function<B(const A&)>;

// A helper struct as template function can't get partial specialization
template <typename... Funcs>
struct ComposerHelper;

// Base case: a single arrow
template <typename A, typename B>
struct ComposerHelper<Arrow<A, B>>
{
    static Arrow<A, B> compose(const Arrow<A, B>& arr)
    {
        return arr;
    }
};

// Tail case: more arrows
template <typename A, typename B, typename... Tail>
struct ComposerHelper<Arrow<A, B>, Tail...>
{
    // hard to know the exact return type here. Let the compiler figure out
    static auto compose(const Arrow<A, B>& arr, const Tail&... tail)
    -> decltype(arr * ComposerHelper<Tail...>::compose(tail...))
    {
        return arr * ComposerHelper<Tail...>::compose(tail...);
    }
};

// A simple function to call our helper struct
// again, hard to write the return type
template <typename... Funcs>
auto compose(const Funcs&... funcs)
-> decltype(ComposerHelper<Funcs...>::compose(funcs...))
{
    return ComposerHelper<Funcs...>::compose(funcs...);
}

using namespace std;

int main()
{
    Arrow<int, float> plusHalf([](int i){return i + 0.5;});
    Arrow<float, string> toString([](float f){return to_string(f);});
    Arrow<string, int> firstDigit([](const string& s){return s[0]-'0';});

    auto composed = compose(plusHalf, toString, firstDigit);
    // composed is Arrow<int, int>

    std::cout << composed(61) << std::endl; // "6"

    //auto badComposed = compose(toString, plusHalf); // compile time error
}

Answer 2

Though the output is exactly the same, here I feel Arrow should be its own class so that its domain and range can be stored as types. Now return types are known without having the compiler deduce them.

#include <functional>
#include <iostream>
#include <string>
#include <sstream>

// Revised as a class with types domain = A and range = B.  It still acts as a std::function<B(const A&)>, through its operator()(const A&) and its data member 'arrow'.
template <typename A, typename B>
class Arrow {
    const std::function<B(const A&)> arrow;
public:
    Arrow (const std::function<B(const A&)>& a) : arrow(a) {}
    B operator()(const A& a) const {return arrow(a);}
    using domain = A;
    using range = B;
};

// The overload * for the composition of two Arrows in Arrow's new form:
template <typename A, typename B, typename C>
Arrow<A,C> operator * (const Arrow<A,B>& arrow_ab, const Arrow<B,C>& arrow_bc) {
    return Arrow<A,C>([arrow_ab, arrow_bc](const A& a)->C {return arrow_bc(arrow_ab(a));});
}

template <typename First, typename... Rest> struct LastType : LastType<Rest...> {};
template <typename T> struct Identity { using type = T; };
template <typename Last> struct LastType<Last> : Identity<Last> {};

template <typename...> struct ComposeArrows;

template <typename First, typename... Rest>
struct ComposeArrows<First, Rest...> : ComposeArrows<Rest...> {
    using last_arrow = typename LastType<Rest...>::type;
    using composed_arrow = Arrow<typename First::domain, typename last_arrow::range>;
    composed_arrow operator()(const First& first, const Rest&... rest) {
        return first * ComposeArrows<Rest...>::operator()(rest...);
    }
};

template <typename Last>
struct ComposeArrows<Last> {
    Last operator()(const Last& arrow) {return arrow;}
};

template <typename... Arrows>
typename ComposeArrows<Arrows...>::composed_arrow composeArrows (const Arrows&... arrows) {
    return ComposeArrows<Arrows...>()(arrows...);
}

// ------------ Testing ------------
template <typename T>
std::string toString (const T& t) {
    std::ostringstream os;
    os << t;
    return os.str();
}

int main() {
    Arrow<int, double> plusHalf ([](int i){return i + 0.5;});
    Arrow<double, std::string> doubleToString ([](float f){return toString(f) + " is the result";});
    Arrow<std::string, char> lastCharacter ([](const std::string& str)->int {show(str)  return str.back();});

    const Arrow<int, char> composed = composeArrows (plusHalf, doubleToString, lastCharacter);
    std::cout << composed(2) << std::endl; // "t"
    std::cout << composeArrows (plusHalf, doubleToString, lastCharacter)(2) << std::endl; // "t"
}

Here is another solution with a slightly different syntax:

#include <iostream>
#include <functional>
#include <tuple>
#include <string>

template <typename A, typename B>
struct Arrow {
    using domain = const A&;
    using range = B;
    using Function = std::function<range(domain)>;
    const Function& function;
    Arrow (const Function& f) : function(f) {}
    range operator()(domain x) const {return function(x);}
};

template <typename... Ts>
struct LastType {
    using Tuple = std::tuple<Ts...>;
    using type = typename std::tuple_element<std::tuple_size<Tuple>::value - 1, Tuple>::type;
};

template <typename Arrow1, typename Arrow2>
typename Arrow2::range compose (const typename Arrow1::domain& x, const Arrow2& arrow2, const Arrow1& arrow1) {
    return arrow2(arrow1(x));
}

template <typename Arrow, typename... Rest>
auto compose (const typename LastType<Rest...>::type::domain& x, const Arrow& arrow, const Rest&... rest) {
    return arrow(compose(x, rest...));
}

int main() {
    Arrow<std::string, int> f([](const std::string& s) {return s.length();});
    Arrow<int, double> g([](int x) {return x + 0.5;});
    Arrow<double, int> h([](double d) {return int(d+1);});
    std::cout << compose("hello", g, f) << '\n';  // 5.5
    std::cout << compose("hello", h, g, f) << '\n';  // 6
}