Functional composition with variadic templates

Question

My goal is to get composition of functions working with this exact syntax:

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

By changing the syntax slightly so that the "hello" argument goes first, I have it working easily with the following code:

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

template <typename D, typename R>
struct Function {
    using domain = const D&;
    using range = R;
    using function = std::function<range(domain)>;
    const function& f;
    Function (const function& f) : f(f) {}
    range operator()(domain x) const {return f(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 F, typename G>
typename G::range compose (const typename F::domain& x, const G& g, const F& f) {
    return g(f(x));
}

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

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

Having done that, I thought it would be a trivial task to adapt the above code so that I get the exact syntax I want (i.e. with "hello" being at the end of the list), but it is turning more difficult than I thought. I attempted the following, which does not compile:

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

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

template <typename F, typename G>
typename G::range compose (const G& g, const F& f, const typename F::domain& x) {
    return g(f(x));
}

template <typename F, typename... Rest>
auto compose (const F& f, const Rest&... rest) {
    return f(compose(rest...));
}

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

And I don't know how to fix it. Can anybody help me fix this?

A new idea I've come up with is to define compose_, which will reorder the arguments of args... (by some std::tuple manipulation) so that the first element goes last and then passing that argument pack to compose. This looks very messy though, and even if it works, there must be a more direct (and shorter) solution.

Answer 1

It looks like the following also works:

template <typename T>
const T& compose (const T& t) {
    return t;
}

template <typename F, typename... Rest>
typename F::range compose(const F& f, Rest... rest) {
    return f(compose(rest...));
}

Answer 2

what about in this way?

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

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

template <typename F, typename X = typename F::domain>
typename F::range compose (const F& f, const X & x) {
    return f(x);
}

template <typename F, typename... Rest>
typename F::range  compose (const F& f, const Rest&... rest) {
    return f(compose(rest...));
}

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

You can use auto for the returning type of compose() only in c++14 (if I'm not wrong).

Your version doesn't compile because your variadic version of compose() uses N variadic types and N arguments when the final (not varidic) uses 2 types and 3 arguments. In other words, the variadic version lost the final argument. Your version doesn't compile because the final (not variadic version) is never used: the compiler chooses the variadic version. Adding typename X = typename F::domain (and changing const typename F::domain& with const X&) the final version is preferred and your code should compile (with c++14, at least) [corrected by Piotr Skotnicki; thanks]

p.s.: sorry for my bad English.