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.