Re: Is this portable? [static pointer casts, char* arithmetic]

On 14 Apr., 19:27, SG <s.gesem...@gmail.com> wrote:


I should add that I expected such a conversion not to create a new
copy (of the "pointee") but just to "work" polymorphically (which is
probably too much to ask for).

A simple example would look like this:

   void test_string()
   {
      cow<string> x = string("hello");
      cow<string> y = x;
      cout << "*x --> " << *x << endl;
      cout << "*y --> " << *y << endl;
      cout << "shared = " << (&*x == &*y) << endl;
      x.wref() += "123"; // "wref" for write access ...
      cout << "*x --> " << *x << endl;
      cout << "*y --> " << *y << endl;
      cout << "shared = " << (&*x == &*y) << endl;
   }

which is supposed to output:

   *x --> hello
   *y --> hello
   shared = 1
   *x --> hello123
   *y --> hello
   shared = 0

Without the polymorphism requirement this would be fairly easy to
achieve. But I also tried to make this code:

   struct base
   {
   private:
      virtual void greet_once() const
      { cout << "hello from base\n"; }
   public:
      void greet() const
      { for (int k=0; k<repeat; ++k) greet_once(); }
      int repeat;
   };

   struct derived : base
   {
   private:
      void greet_once() const
      { cout << "hello from derived\n"; }
   public:
      ~derived()
      { cout << "this is ~derived speaking\n"; }
   };

   void test_polymorphism()
   {
      cow<base> x = derived();
      x.wref().repeat = 3;
      x->greet();
   }

to output the following text:

   this is ~derived speaking
   hello from derived
   hello from derived
   hello from derived
   this is ~derived speaking

where the first "~derived" message is due to the temporary in the copy-
initialization and the 2nd one comes from the copy of derived hidden
behind a type-agnostic abstract wrapper.

Cheers!
SG