[Boost-users] Re: Placeholder Arguments

From: David Abrahams (dave_at_[hidden])
Date: 2005-04-27 05:28:25

Joe McCay <joemccay_at_[hidden]> writes:

> David Abrahams <dave_at_[hidden]> writes:
>> Are you asking about runtime lambda expressions as in the Boost Lambda
>> library, or compile-time lambda expressions as in MPL?
>> 
>> --
>> Dave Abrahams
>> Boost Consulting
>> www.boost-consulting.com
>
>    I was talking about MPL because that is the only one I have studied
> at the moment.  What is the difference between the two outside of the
> obvious runtime/compile-time difference?  

Well, the syntax.  MPL's compile-time lambdas never use parens as in
the _1(5) you cited.

> I was trying to figure out how a particular placeholder gets its
> arguments to choose from when deciding the argument (argument 1 for
> _1 etc.).

It's just in the definition of the placeholders 

  template <int> struct arg;
  typedef arg<1> _1;
  typedef arg<2> _2;
  …
  template <> struct arg<N>
  {
      template <
          class A1 = void_
        , class A2 = void_
          …
        , class AM = void_
      >
      struct apply
      {
          typedef AN type;
      };
  };

and the way arguments are passed through from apply_wrap.

    apply_wrap<_2, int, int*, int**, int***>::type => int*

Maybe this simplified definition of lambda would help:

  template <class T> 
  struct lambda { typedef T type; };

  template <  template <class> class F, class A1  >
  struct lambda< F<A1> > 
    : if_<
          is_placeholder_expression< F<A1> >
        , bind< quote1<F>, typename lambda<A1>::type >
        , F<A1>
      >
  {};

  template <  template <class,class> class F, class A1, class A2  >
  struct lambda< F<A1,A2> > 
    : if_<
          is_placeholder_expression< F<A1,A2> >
        , bind< 
              quote2<F>
            , typename lambda<A1>::type, typename lambda<A2>::type 
          >
        , F<A1,A2>
      >
  {};

  //...

-- 
Dave Abrahams
Boost Consulting
www.boost-consulting.com