**not**be explaining these features, or how and why they work as that is too much for one blog post to accomplish, but I will be explaining how they work in context.

Note that if one understood the article up to the definition of

*ConstructT*, then feel free to skip to that section.

**Update:**I have attempted to improve the original article with some of this one, so it may be a little redundant.

#### Binary

Binary defines member functions for partially applying the first and last arguments. I wrote about this back in September.template< class F, class X > struct Part { F f; X x; template< class _F, class _X > constexpr Part( _F&& f, _X&& x ) : f(forward<_F>(f)), x(forward<_X>(x)) { } template< class ... Xs > constexpr auto operator () ( Xs&& ...xs ) -> decltype( f(x,declval<Xs>()...) ) { return f( x, forward<Xs>(xs)... ); } }; template< class F, class X > struct RPart { F f; X x; template< class _F, class _X > constexpr RPart( _F&& f, _X&& x ) : f( forward<_F>(f) ), x( forward<_X>(x) ) { } template< class ...Y > constexpr decltype( f(declval<Y>()..., x) ) operator() ( Y&& ...y ) { return f( forward<Y>(y)..., x ); } }; template< class F > struct Binary { template< class X > constexpr Part<F,X> operator () ( X x ) { return Part<F,X>( F(), move(x) ); } template< class X > constexpr RPart<F,X> with( X x ) { return RPart<F,X>( F(), move(x) ); } };

So, given some type,

*F*, inherited from

*Binary<F>*,

*F*is defined as taking one argument (partial application), and calling the member function

*with*(reverse-partial application).

*F*must be a type that defines a two-argument

*operator()*overload.

We define a derivation like so:

constexpr struct Add : public Binary<Add> { using Binary<Add>::operator(); template< class X, class Y > constexpr auto operator () ( X&& x, Y&& y ) -> decltype( std::declval<X>() + std::declval<Y>() ) { return std::forward<X>(x) + std::forward<Y>(y); } } add{};

*Add*, is not a function in the C sense, it's a function type.

*add*is an instance of

*Add*and is a function object.

*Add*does not inherit

*Binary's operator()*overload by default, so we explicitly tell it to by saying "

*using Binary<Add>::operator()*".

*Binary's*

*with*requires no work to inherit.

*auto inc = add(1); // Calls Binary<Add>::operator(); returns Part<Add,int>.*

*auto dec = add.with(-1); // Calls Binary<Add>::with; returns RPart<Add,int>.*

*int two = inc(1); // Calls Part<Add,int>::operator(); returns add(1,1).*

*int one = dec(two); // returns add(2,-1).*

*add(1,2) -- Calls Add::operator(); returns int(3).*

####
*Chainable*.

template< class F > struct Chainable : Binary<F> { using Binary<F>::operator(); template< class X, class Y > using R = typename std::result_of< F(X,Y) >::type; // Three arguments: unroll. template< class X, class Y, class Z > constexpr auto operator () ( X&& x, Y&& y, Z&& z ) -> R< R<X,Y>, Z > { return F()( F()( std::forward<X>(x), std::forward<Y>(y) ), std::forward<Z>(z) ); } template< class X, class Y, class ...Z > using Unroll = typename std::result_of < Chainable<F>( typename std::result_of<F(X,Y)>, Z... ) >::type; // Any more? recurse. template< class X, class Y, class Z, class H, class ...J > constexpr auto operator () ( X&& x, Y&& y, Z&& z, H&& h, J&& ...j ) -> Unroll<X,Y,Z,H,J...> { // Notice how (*this) always gets applied at LEAST three arguments. return (*this)( F()( std::forward<X>(x), std::forward<Y>(y) ), std::forward<Z>(z), std::forward<H>(h), std::forward<J>(j)... ); } };

*Chainable*works in much the same way as

*Binary*does, except that it extends

*F*to take any arbitrary number of arguments (except for zero, but that'd be pretty useless anyway). Redefining

*Add*to be

*Chainable*is not hard.

constexpr struct Add : public Chainable<Add> { using Chainable<Add>::operator(); template< class X, class Y > constexpr auto operator () ( X&& x, Y&& y ) -> decltype( std::declval<X>() + std::declval<Y>() ) { return std::forward<X>(x) + std::forward<Y>(y); } } add{};

Only two lines have changed, however

*add*'s behaviour has changed dramatically.

*int x = add(1,2,3); // Calls Chainable<Add>::operator(X,Y,Z); returns (1+2)+3.*

*int y = add(1,2,3,4,5,6); // Calls Chainable<Add>::operator(X,Y,Z,H,J...).*

*auto inc = add(1); // Still calls Binary<Add>::operator().*

*The three arg version of*

*Chainable<Add>*calls

*add( add(x,y), z )*, while the variadic version calls

*itself( add(x,y), z, h, j... )*. It reduces the number of arguments to process by one, being supplied at least four. It is therefore never the base case and always ends by calling its three-arg version.

####
*ConstructT*.

template< template<class...> class T > struct ConstructT { template< class ...X, class R = T< typename std::decay<X>::type... > > constexpr R operator () ( X&& ...x ) { return R( forward<X>(x)... ); } };

The first template argument may be confusing to those who have not seen them before. We could write

*template<class> class T*, and that would expect a

*T*that took one type parameter. It may actually take one, zero, or several.

*template<class...> class T*is the generic way to say "we know

*T*takes type parameters, but we don't know how many (and it doesn't yet matter)". It might be

*std::vector*, which can take two; the value type and allocator. It might be

*std::set*, which can take three. (Though, neither

*vector*nor

*set*would work for

*ConstructT*.)

We deduce the return type using the default template parameter,

*R*.

*std::decay*transforms the type such that if we pass in an

*int&*, we get an

*int*back.

*typename std::decay<const int&>::type = int*

*typename std::decay<int&&>::type = int*

*typename std::decay<int>::type = int*

So, if

*T=std::pair*and we pass in an

*int&*and

*std::string&&*,

*R = std::pair<int,std::string>*.

*ConstructT*perfectly forwards the arguments to

*T*'s constructor, but decays the types to ensure that it holds the actual values.

Just the same as with

*add*and

*Add*, we must create an instance of

*ConstructT*to call it.

*constexpr auto make_pair = ConstructT<std::pair>();*

This version of

*make_pair*behaves just like

*std::make_pair*. Its type is equivalent to this:

template< struct ConstructPair { template< class ...X, class R = std::pair< typename std::decay<X>::type... > > constexpr R operator () ( X&& ...x ) { return R( forward<X>(x)... ); } };

#### Conclusion.

C++11 is very big and no one feature is incredibly useful, but in conjunction they build up a powerful synergy. If these features had been a part of the language to begin with, there is no question that they would be considered required reading, just as much as*std::vector*and the

*<algorithm>*library are today. They solve problems that have led to years frustration in developers. Perhaps one of the reasons many hate C++ is because of the lacking of many of these features. Becoming fluent in them makes C++ a much nicer language to speak and communicate with.