Re: [Boost-users] Python 'in' keyword-like in C++

22 Sep 2005

      Tzu-Chien Chiu wrote:
...
Any boost library can make me write the expression like
int x = 3;
 if ( x in list(3, 5, 6, 7) ) {
  ...
 }
I don't know about the boost libraries, but I quickly sketched out 
something you might be interested in, based on something I saw on one of 
the news groups. All you have to do is to write a binary predicate 
functor with bool operator()(int,<whatever your list type is>). It's 
untested, but a quick syntax check with g++ 3.4.2 works fine.

Usage will be
if (x <in> list(3,5,6,7) ) { ... }

I hope that's more or less what you're after; I don't know if you want 
that list type/function as well.

The code itself is attached. Be sure to fix the template parameter in 
the unnamed namespace..

David Hall

template <class F>
struct operator_t
{
 operator_t(F f=F()): f(f) {}
 F f;
};

// Make a binary predicate functor that takes two arguments that iterates over your list...
namespace
{
operator_t<YOU FILL IN HERE> in;
}

namespace detail
{
template 
struct operator_t_bind
{
operator_t_bind(const T & x,F & f):lhs(x),f(f) {}
const T & lhs;
const F & f;
};
}

template 
detail::operator_t_bind operator<(const T & lhs,operator_t<F> & op)
{
  return detail::operator_t_bind(lhs,op.f);
}

template 
bool operator>(const detail::operator_t_bind & binder,const RHS & rhs)
{
 return binder.f(binder.lhs,rhs);
}