Re: Constructor syntax woes.

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Wed, 16 Jul 2008 02:24:06 +0200
Message-ID:
<kd6dnWBel586ouDVnZ2dnUVZ_t_inZ2d@posted.comnet>
* willo@cynd.net:

Below is a condensed version of some code I'm having difficulty with:

===============================================================

#include <cstddef> // size_t


Well, with the current standard formally this only gives you std::size_t, but
all or nearly all compilers also place it in the global namespace. With C++0x
it's allowed to also give you size_t in global namespace. Because of that it's
now, IMHO, misleading to use the <c...> headers; just use <stddef.h>.

#include <functional> // binary_function

// the input parameter type to a class constructor.
template< typename T >
struct ABinaryFunction : public std::binary_function< T, size_t,
double >
{
  typename ABinaryFunction::result_type
  operator()( typename ABinaryFunction< T >::first_argument_type lhs,
              typename ABinaryFunction< T >::second_argument_type
rhs )
  {
    return ABinaryFunction< T >::result_type();
  }
};


OK, except you'll probably want to add a 'const' for the operator() (presumably
it's not changing the functor object, but just computing something).

template< typename T >
class AClass
{
  public:

  // constructor takes one specialized binary_function type parameter.
  AClass( std::binary_function< T, size_t, double > function )
  {
  }


If you want to somehow retain the functor object then you need to use some other
kind of argument, possibly a templated constructor. As it is your actual
argument will be sliced to std::binary_function. Not much you can do with a pure
std::binary_function!

  // arbitrary method to test instantiation.
  bool True()
  {
    return true;
  }
};

int main()
{
  // this won't create a class of type "AClass".
  AClass< int > instance_a( ABinaryFunction< int >() );


This is an example of what's been called "the most vexing parse" in C++. The
rule is that if the compiler can treat a declaration as a function declaration,
it will. And here it can, so it does.

You can fiddle with extra parentheses and the like, to convince the compiler
that that argument can't possibly be a type, but that yields unclear code.

So instead just do

   ABinaryFunction<int> foo;
   AClass<int> instance_a( foo );

Or you can introduce a factory function for your binary functors,

   template< typename T >
   ABinaryFunction<T> aBinaryFunction() { return ABinaryFunction<T>(); }

and then in main you can declare

   AClass<int> instance_a( aBinaryFunction<int>() );

Anyway, as noted earlier, this actual argument will be sliced.

  // this will not compile.
  bool bool_a = instance_a.True();

  // this will create a class of type "AClass", along with an unwanted
binary_function_b.
  ABinaryFunction< int > binary_function_b;
  AClass< int > instance_b( binary_function_b );
  // this will compile.
  bool bool_b = instance_b.True();


Yes, that's OK.

  return 0;
}


Cheers, & hth.,

- Alf

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