Re: The joy (?) of C++98 argument forwarding

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Tue, 04 May 2010 15:50:18 +0200
Message-ID:
<hrp8qm$n69$1@news.eternal-september.org>
On 04.05.2010 15:01, * Alf P. Steinbach:

On 04.05.2010 14:56, * DeMarcus:

Alf P. Steinbach wrote:

Consider ...


[snip]

Shared< Foo > r4( newObject(), args( 42, "blah blah" ) );

[snip]

Is the notation in the 'test' routine OK, or would it be better (in
some sense) with some macro, e.g. like

Shared< Foo > r( CPPX_NEW_SHARED( 42, "blah blah" ) );

?


At least consider CPPX_NEW_SHARED_FOO( 42, "blah blah" ). Otherwise the
macro name would be as non-intuitive as

CPPX_NEW_SHARED( PI, 21.45, time() )


Well for the example above the type is implicit in the declaration, so
it would be redundant to repeat it there.

On the other hand you have a point -- I didn't think of it -- for an
actual argument in a routine call.

On the third hand, we usually don't specify arguments there for value
arguments.


Keyboard gremlin was active, I meant, "we usually don't specify the type for an
actual argument for a value argument".

Can't you do the wrapper with a template instead of a macro?


Not sure what you mean; the Shared shared pointer type is templated, as
the example usage code shows.


Oh, I think I see what you meant now. You probably thought that the newObject()
call produces a Foo*.

It doesn't, it's just a placeholder thing, of fixed type.

In

     Shared< Foo > r4( newObject(), args( 42, "blah blah" ) );

the arguments are just arguments, and it's the Shared pointer constructor that
does the new'ing.

Happily the Shared definition is not more code than I can show in Usenet article[1].

The constructor invoked by the above is the second one:

<code>
namespace progrock{ namespace cppx{

     enum NewObject {};
     inline NewObject newObject() { return NewObject(); }

     template< typename Type >
     class Shared
         : protected detail::SharedBase< Type >
     {
     private:
         typedef detail::SharedBase< Type > Base;

     public:
         typedef Type Object;

         Shared( NewObject )
             : Base( detail::newRefCounted< Object >( EmptyArgPack() ) )
         {}

         template< class ArgPack >
         Shared( NewObject, ArgPack const& args )
             : Base( detail::newRefCounted< Object >( args ) )
         {}

         template< class OtherType >
         Shared( std::auto_ptr< OtherType > p )
             : Base(
                 detail::newRefCounted< detail::RawPtrHolder< OtherType > >(
args( p.get() ) ),
                 p.get()
                 )
         {
             p.release();
         }

         template< class OtherType >
         Shared( Shared< OtherType > const& other )
             : Base( other )
         {}

         void swapWith( Shared& other )
         {
             Base::swapWith( other );
         }

         Shared& operator=( Shared other )
         {
             swapWith( other ); return *this;
         }

         Object* rawPtr() const { return myObject; }
         Object& rawRef() const { return *rawPtr() }
         Object& operator*() const { return rawPtr(); }
         Object* operator->() const { return rawPtr(); }
     };
} } // namespace progrock::cppx

template< class Type >
inline CPPX_IMPLEMENT_SWAP( progrock::cppx::Shared< Type > )
</code>

The question still stands, is it OK with notation like

     Shared< Foo > r4( newObject(), args( 42, "blah blah" ) );

or would it be better in some sense with some macro, e.g. like

     Shared< Foo > r( CPPX_NEW_SHARED( 42, "blah blah" ) );

?

Cheers,

- Alf

Notes:
[1] There is a subtlety here regarding copy construction. The templated
constructor that looks like a copy constructor isn't one. The compiler generated
copy constructor invokes the one of the base class.

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