How can currying be done in C++?

Question 1

How can currying be done in C++?

c++ stl functional-programming currying binders

yesraaj · Sep 30, 2008 · Viewed 27.2k times · Source

Answer

Answer

1. What is currying?

Currying simply means a transformation of a function of several arguments to a function of a single argument. This is most easily illustrated using an example:

Take a function f that accepts three arguments:

int
f(int a,std::string b,float c)
{
    // do something with a, b, and c
    return 0;
}

If we want to call f, we have to provide all of its arguments f(1,"some string",19.7f).

Then a curried version of f, let's call it curried_f=curry(f) only expects a single argument, that corresponds to the first argument of f, namely the argument a. Additionally, f(1,"some string",19.7f) can also be written using the curried version as curried_f(1)("some string")(19.7f). The return value of curried_f(1) on the other hand is just another function, that handles the next argument of f. In the end, we end up with a function or callable curried_f that fulfills the following equality:

curried_f(first_arg)(second_arg)...(last_arg) == f(first_arg,second_arg,...,last_arg).

2. How can currying be achieved in C++?

The following is a little bit more complicated, but works very well for me (using c++11)... It also allows currying of arbitrary degree like so: auto curried=curry(f)(arg1)(arg2)(arg3) and later auto result=curried(arg4)(arg5). Here it goes:

#include <functional>

namespace _dtl {

    template <typename FUNCTION> struct
    _curry;

    // specialization for functions with a single argument
    template <typename R,typename T> struct
    _curry<std::function<R(T)>> {
        using
        type = std::function<R(T)>;
        
        const type
        result;
        
        _curry(type fun) : result(fun) {}
        
    };

    // recursive specialization for functions with more arguments
    template <typename R,typename T,typename...Ts> struct
    _curry<std::function<R(T,Ts...)>> {
        using
        remaining_type = typename _curry<std::function<R(Ts...)> >::type;
        
        using
        type = std::function<remaining_type(T)>;
        
        const type
        result;
        
        _curry(std::function<R(T,Ts...)> fun)
        : result (
            [=](const T& t) {
                return _curry<std::function<R(Ts...)>>(
                    [=](const Ts&...ts){ 
                        return fun(t, ts...); 
                    }
                ).result;
            }
        ) {}
    };
}

template <typename R,typename...Ts> auto
curry(const std::function<R(Ts...)> fun)
-> typename _dtl::_curry<std::function<R(Ts...)>>::type
{
    return _dtl::_curry<std::function<R(Ts...)>>(fun).result;
}

template <typename R,typename...Ts> auto
curry(R(* const fun)(Ts...))
-> typename _dtl::_curry<std::function<R(Ts...)>>::type
{
    return _dtl::_curry<std::function<R(Ts...)>>(fun).result;
}

#include <iostream>

void 
f(std::string a,std::string b,std::string c)
{
    std::cout << a << b << c;
}

int 
main() {
    curry(f)("Hello ")("functional ")("world!");
    return 0;
}

View output

OK, as Samer commented, I should add some explanations as to how this works. The actual implementation is done in the _dtl::_curry, while the template functions curry are only convenience wrappers. The implementation is recursive over the arguments of the std::function template argument FUNCTION.

For a function with only a single argument, the result is identical to the original function.

        _curry(std::function<R(T,Ts...)> fun)
        : result (
            [=](const T& t) {
                return _curry<std::function<R(Ts...)>>(
                    [=](const Ts&...ts){ 
                        return fun(t, ts...); 
                    }
                ).result;
            }
        ) {}

Here the tricky thing: For a function with more arguments, we return a lambda whose argument is bound to the first argument to the call to fun. Finally, the remaining currying for the remaining N-1 arguments is delegated to the implementation of _curry<Ts...> with one less template argument.

Update for c++14 / 17:

A new idea to approach the problem of currying just came to me... With the introduction of if constexpr into c++17 (and with the help of void_t to determine if a function is fully curried), things seem to get a lot easier:

template< class, class = std::void_t<> > struct 
needs_unapply : std::true_type { };
 
template< class T > struct 
needs_unapply<T, std::void_t<decltype(std::declval<T>()())>> : std::false_type { };

template <typename F> auto
curry(F&& f) {
  /// Check if f() is a valid function call. If not we need 
  /// to curry at least one argument:
  if constexpr (needs_unapply<decltype(f)>::value) {
       return [=](auto&& x) {
            return curry(
                [=](auto&&...xs) -> decltype(f(x,xs...)) {
                    return f(x,xs...);
                }
            );
        };    
  }
  else {  
    /// If 'f()' is a valid call, just call it, we are done.
    return f();
  }
}

int 
main()
{
  auto f = [](auto a, auto b, auto c, auto d) {
    return a  * b * c * d;
  };
  
  return curry(f)(1)(2)(3)(4);
}

See code in action on here. With a similar approach, here is how to curry functions with arbitrary number of arguments.

The same idea seems to work out also in C++14, if we exchange the constexpr if with a template selection depending on the test needs_unapply<decltype(f)>::value: