How can currying be done in C++?
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:
template <typename F> auto
curry(F&& f);
template <bool> struct
curry_on;
template <> struct
curry_on<false> {
template <typename F> static auto
apply(F&& f) {
return f();
}
};
template <> struct
curry_on<true> {
template <typename F> static auto
apply(F&& f) {
return [=](auto&& x) {
return curry(
[=](auto&&...xs) -> decltype(f(x,xs...)) {
return f(x,xs...);
}
);
};
}
};
template <typename F> auto
curry(F&& f) {
return curry_on<needs_unapply<decltype(f)>::value>::template apply(f);
}
Simplifying Gregg's example, using tr1:
#include <functional>
using namespace std;
using namespace std::tr1;
using namespace std::tr1::placeholders;
int f(int, int);
..
int main(){
function<int(int)> g = bind(f, _1, 5); // g(x) == f(x, 5)
function<int(int)> h = bind(f, 2, _1); // h(x) == f(2, x)
function<int(int,int)> j = bind(g, _2); // j(x,y) == g(y)
}
Tr1 functional components allow you to write rich functional-style code in C++. As well, C++0x will allow for in-line lambda functions to do this as well:
int f(int, int);
..
int main(){
auto g = [](int x){ return f(x,5); }; // g(x) == f(x, 5)
auto h = [](int x){ return f(2,x); }; // h(x) == f(2, x)
auto j = [](int x, int y){ return g(y); }; // j(x,y) == g(y)
}
And while C++ doesn't provide the rich side-effect analysis that some functional-oriented programming languages perform, const analysis and C++0x lambda syntax can help:
struct foo{
int x;
int operator()(int y) const {
x = 42; // error! const function can't modify members
}
};
..
int main(){
int x;
auto f = [](int y){ x = 42; }; // error! lambdas don't capture by default.
}
Hope that helps.
In short, currying takes a function f(x, y) and given a fixed Y, gives a new function g(x) where
g(x) == f(x, Y)
This new function may be called in situations where only one argument is supplied, and passes the call on to the original f function with the fixed Y argument.
The binders in the STL allow you to do this for C++ functions. For example:
#include <functional>
#include <iostream>
#include <vector>
using namespace std;
// declare a binary function object
class adder: public binary_function<int, int, int> {
public:
int operator()(int x, int y) const
{
return x + y;
}
};
int main()
{
// initialise some sample data
vector<int> a, b;
a.push_back(1);
a.push_back(2);
a.push_back(3);
// here we declare a function object f and try it out
adder f;
cout << "f(2, 3) = " << f(2, 3) << endl;
// transform() expects a function with one argument, so we use
// bind2nd to make a new function based on f, that takes one
// argument and adds 5 to it
transform(a.begin(), a.end(), back_inserter(b), bind2nd(f, 5));
// output b to see what we got
cout << "b = [" << endl;
for (vector<int>::iterator i = b.begin(); i != b.end(); ++i) {
cout << " " << *i << endl;
}
cout << "]" << endl;
return 0;
}