使用模板元编程在C ++中实现“深层”功能

Question

I just came up with an (yet another!) implementation of function currying in C++ using template metaprogramming. (I am almost sure other implementations are better / more complete than mine, but I am doing this for learning purposes, a case in which I think reinventing the wheel is justified.)

My funcion currying implementation, test case included, is the following one:

#include <iostream>
#include <functional>

template <typename> class curry;

template <typename _Res>
class curry< _Res() >
{
  public:
    typedef std::function< _Res() > _Fun;
    typedef _Res _Ret;

  private:
    _Fun _fun;

  public:
    explicit curry (_Fun fun)
    : _fun(fun) { }

    operator _Ret ()
    { return _fun(); }
};

template <typename _Res, typename _Arg, typename... _Args>
class curry< _Res(_Arg, _Args...) >
{
  public:
    typedef std::function< _Res(_Arg, _Args...) > _Fun;
    typedef curry< _Res(_Args...) > _Ret;

  private:
    class apply
    {
      private:
        _Fun _fun;
        _Arg _arg;

      public:
        apply (_Fun fun, _Arg arg) 
        : _fun(fun), _arg(arg) { }

        _Res operator() (_Args... args)
        { return _fun(_arg, args...); }
    };

  private:
    _Fun _fun;

  public:
    explicit curry (_Fun fun)
    : _fun(fun) { }

    _Ret operator() (_Arg arg)
    { return _Ret(apply(_fun, arg)); }
};

int main ()
{
  auto plus_xy = curry<int(int,int)>(std::plus<int>());
  auto plus_2x = plus_xy(2);
  auto plus_24 = plus_2x(4);
  std::cout << plus_24 << std::endl;

  return 0;
}

This function currying implementation is "shallow", in the following sense: If the original std::function 's signature is...

(arg1, arg2, arg3...) -> res

Then the curried function's signature is...

arg1 -> arg2 -> arg3 -> ... -> res

However, if any of the arguments or the return type themselves can be curried, they do not get curried. For example, if the original std::function 's signature is...

(((arg1, arg2) -> tmp), arg3) -> res

Then the curried function's signature will be...

((arg1, arg2) -> tmp) -> arg3 -> res

Instead of...

(arg1 -> arg2 -> tmp) -> arg3 -> res

Which is what I want. So I would like to have a "deep" currying implementation. Does anyone know how I could write it?

---

@vhallac:

This is the kind of function that should be passed to the constructor of curry<int(int(int,int),int)> :

int test(std::function<int(int,int)> f, int x)
{ return f(3, 4) * x; }

Then one should be able to do the following:

auto func_xy = curry<int(int(int,int),int)>(test);
auto plus_xy = curry<int(int,int)>(std::plus<int>());
auto func_px = func_xy(plus_xy);
auto func_p5 = func_px(5);
std::cout << func_p5 << std::endl;

Answer 1

I have implemented a cheating version of a decurry class to demonstrate how you would go about implementing the specialization. The version is cheating, because it gets declared as a friend of curry<T> , and accesses the internal _fun to convert a curried version of a function back to original. It should be possible to write a generic one, but I didn't want to spend more time on it.

The decurry implementation is:

template <typename _Res, typename... _Args>
class decurry< curry<_Res(_Args...)> > {
public:
    typedef curry<_Res(_Args...)> _Curried;
    typedef typename curry<_Res(_Args...)>::_Fun _Raw;

    decurry(_Curried fn): _fn(fn) {}

    _Res operator() (_Args... rest) {
        return _fn._fun(rest...);
    }
private:
    _Curried _fn;
};

And it requires the line:

friend class decurry< curry<_Res(_Arg, _Args...)> >;

inside class curry< _Res(_Arg, _Args...) > to give our class access to curry<T>._fun .

Now, the specialization can be written as:

template <typename _Res, typename _Res2, typename... _Args2, typename... _Args>
class curry< _Res(_Res2(_Args2...), _Args...) >
{
public:
    typedef curry< _Res2(_Args2...) > _Arg;
    typedef std::function< _Res2(_Args2...) > _RawFun;
    typedef std::function< _Res(_RawFun, _Args...) > _Fun;
    typedef curry< _Res(_Args...) > _Ret;

private:
    class apply
    {
    private:
        _Fun _fun;
        _RawFun _arg;

    public:
        apply (_Fun fun, _RawFun arg)
            : _fun(fun), _arg(arg) { }

        _Res operator() (_Args... args)
        { return _fun(_arg, args...); }
    };

private:
    _Fun _fun;

public:
    explicit curry (_Fun fun)
        : _fun(fun) { }

    _Ret operator() (_Arg arg)
    { return _Ret(apply(_fun, decurry<_Arg>(arg))); }
};

The test code is a specified in the question:

int test(std::function<int(int,int)> f, int x)
{ return f(3, 4) * x; }

int main ()
{
    auto func_xy = curry<int(int(int,int),int)>(test);
    auto plus_xy = curry<int(int,int)>(std::plus<int>());
    auto func_px = func_xy(plus_xy);
    auto func_p5 = func_px(5);
    std::cout << func_p5 << std::endl;

    return 0;
}

The code output is on Ideone.com again.