TITLE: Delete operator versus operator delete


PROBLEM: cory@shell.portal.com (Coryphaeus - Software)

The ARM seems to be unclear about the order of evaluation
between an operator delete and its destructor.

p 575-576

"Destructors are invoked implicitly ....(3) through use of
the delete operator ..... When invoked by the delete operator
memory is freed by the destructor for the most derived
class of the object using an operator delete."

p 578

"A destructor finds the operator delete() to use for freeing
store using the usual scope rules."


These two seem to contradict.


RESPONSE: rmartin@rcmcon.com (Robert Martin), 23 Sept 94

No, but they are confusing.  There is a difference between "the delete
operator" and "operator delete()".   The "delete operator" is invoked
by the 'delete' keyword.  It then calls the destructor, and then calls
"operator delete".   You see?  There are three (3) different functions
employed here.

There is an interaction between the 'delete operator', the destructor,
and 'operator delete' that is not obvious.   I can psuedo code it as
follows: 

'delete operator'( Object* o)
{
    void* p; // will hold address of buffer to delete;
    size_t s; // will hold size of buffer to delete;
    o->Object::~Object(p,s); // pass in references to these variables
                             // that the destructor will load.
    operator delete(p,s); // Tell operator delete the address and size
                          // of the block to be deleted.
}

This strange piece of pseudo code is trying to make the point that
the 'delete operator' procures both the address and size of the
allocated memory from the destructor of the object.  However, these
values cannot be correct unless the most derived destructor is
executed.  Thus, in the general case, the destructor must be a virtual
function.