[Voted into WP at October 2004 meeting.]

Proposed Resolution (October 2003):

Add the following text at the end of 7.6.2.8 [expr.new] paragraph 19:

If the lookup finds the two-parameter form of a usual deallocation function (6.7.6.5.3 [basic.stc.dynamic.deallocation]), and that function, considered as a placement deallocation function, would have been selected as a match for the allocation function, the program is ill-formed. [Example:

struct S {
  // Placement allocation function:
  static void* operator new(std::size_t, std::size_t);

  // Usual (non-placement) deallocation function:
  static void operator delete(void*, std::size_t);
};

S* p = new (0) S; // ill-formed: non-placement deallocation function matches 
                  // placement allocation function 

--- end example]

Notes from October 2003 meeting:

There was widespread agreement that the compiler shouldn't just silently call the delete with either of the possible values. In the end, we decided it's smarter to issue an error on this case and force the programmer to say what he means.

Mike Miller's analysis of the status quo: 6.7.6.5.3 [basic.stc.dynamic.deallocation] paragraph 2 says that "operator delete(void*, std::size_t)" is a "usual (non-placement) deallocation function" if the class does not declare "operator delete(void*)." 6.7.6.5.2 [basic.stc.dynamic.allocation] does not use the same terminology for allocation functions, but the most reasonable way to understand the uses of the term "placement allocation function" in the Standard is as an allocation function that has more than one parameter and thus can (but need not) be called using the "new-placement" syntax described in 7.6.2.8 [expr.new]. (In considering issue 127, the core group discussed and endorsed the position that, "If a placement allocation function has default arguments for all its parameters except the first, it can be called using non-placement syntax.")

7.6.2.8 [expr.new] paragraph 19 says that any non-placement deallocation function matches a non-placement allocation function, and that a placement deallocation function matches a placement allocation function with the same parameter types after the first -- i.e., a non-placement deallocation function cannot match a placement allocation function. This makes sense, because non-placement ("usual") deallocation functions expect to free memory obtained from the system heap, which might not be the case for storage resulting from calling a placement allocation function.

According to this analysis, the example shows a placement allocation function and a non-placement deallocation function, so the deallocation function should not be invoked at all, and the memory will just leak.

What does this example do?

  #include <stdio.h>
  #include <stdlib.h>

  struct A {
        void* operator new(size_t alloc_size, size_t dummy=0) {
                printf("A::operator new()\n");
                return malloc(alloc_size);
        };

        void operator delete(void* p, size_t s) {
                printf("A::delete %d\n", s);
        };

        A()  {printf("A constructing\n"); throw 2;};

  };

  int
  main() {
    try {
        A* ap = new A;
        delete ap;
    }
    catch(int) {printf("caught\n"); return 1;}
  }

The fundamental issue here is whether the deletion-on-throw is driven by the syntax of the new (placement or non-placement) or by signature matching. If the former, the operator delete would be called with the second argument equal to the size of the class. If the latter, it would be called with the second argument 0.

Core issue 127 (in TC1) dealt with this topic. It removed some wording in 14.3 [except.ctor] paragraph 2 that implied a syntax-based interpretation, leaving wording in 7.6.2.8 [expr.new] paragraph 19 that is signature-based. But there is no accompanying rationale to confirm an explicit choice of the signature-based approach.

EDG and g++ get 0 for the second argument, matching the presumed core issue 127 resolution. But maybe this should be revisited.