Re: How to get arity from lamda expression?

From:

"Peter Dimov" <pdimov@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

21 Jun 2006 05:54:15 -0400

Message-ID:

<1150807993.392483.33790@r2g2000cwb.googlegroups.com>

iwongu wrote:

After I learned some Ruby syntax, I could write the code that print
one's tables in a line like the following.

(2..9).each { |i| (1..9).each { |j| puts "#{i}*#{j}=#{i * j}" } }

So, I thought it's good to have something like this in C++.
The written code is this.

----
using namespace std;
using namespace boost;
using namespace boost::lambda;
using namespace boost::mpl;

class range
{
public:
   range() {
   }
   range(int s, int e) {
     s_.push_back(s);
     e_.push_back(e);
   }

   range& operator()(int s, int e) {
     s_.push_back(s);
     e_.push_back(e);

     return *this;
   }

   template <class F>
   range& operator()(F func) {
     // now this range only support 2-depth loop.
     // but it can be extended.
     typedef typename if_c<F::arity == 2, do_2, do_1>::type do_type;

     do_type()(s_, e_, func);

     return *this;
   }

In principle, you don't need F::arity here; s_.size() tells you what to
do. Relying on arity, even if lambda objects did expose one (they don't
because they have a range of arities) would not allow you to express
the equivalent of

(2..9).each { |i| (1..9).each { |j| puts "#{i}" } }

where the arity of the lambda is one, but it is still used in two
nested loops. (This may or may not be valid Ruby code. :-) )

There's the problem of f( i ) not compiling for cout << _2, though; so
you'll need to drop the single dynamically-dimensioned range class and
return range2 or range<2> from operator()( int, int ).

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