Re: How to get rid of the new-initializer in a new-expression

From:
Jeff Schwab <jeff@schwabcenter.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 21 Feb 2008 22:34:04 CST
Message-ID:
<zq6dnZqTEfaeYyDanZ2dnUVZ_gednZ2d@comcast.com>
Matthias Hofmann wrote:

Hello everyone!

I have written a memory tracker using macros, and everything works fine
unless the new-expression contains a new-initializer. Please take a look at
the following example:

template <class T> T* TrackNew( T* ptr ) { return ptr; }

#define NEW( T ) TrackNew<T>( new T )

struct X
{
     X() {}
     X( int ) {}
};

int main()
{
     // Works fine.
     X* p1 = NEW( X );

     // Fails to compile.
     X* p2 = NEW( X( 2 ) );

     return 0;
}

The second use of the NEW macro expands to:

X* p2 = TrackNew<X( 2 )>( new X( 2 ) );

Obviously, this should be:

X* p2 = TrackNew<X>( new X( 2 ) );

Can anyone please tell me a portable way of getting rid of the
new-initializer in the new-expression? Like in the following pseudo code:

#define NEW( T ) TrackNew<REMOVE_NEW_INITIALIZER( T )>( new T )


Would you consider something like this?

     template<typename T>
     T* New() { return new T; }

     template<typename T, typename U>
     T* New(U const& u) { return new T(u); }

     template<typename T, typename U, typename V>
     T* New(U const& u, V const& v) { return new T(u, v); }

     // ... Up to some large number of arguments.

     struct X {
          X() { }
          X(int) { }
          X(int, double) { }
     };

     int main() {

          X* p1 = New<X>();
          X* p2 = New<X>(2);
          X* p3 = New<X>(2, 3.0);

          // ...

          return 0;
     }

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