Does C++14 require that the delete expression must call `void operator ::delete(void*, std::size_t)` instead of `void ::operator delete(void*)`?

Question

According to this

void operator delete  (void*);                                    (1)   
void operator delete[](void*);                                    (2)   
void operator delete  (void*, const std::nothrow_t&);             (3)   
void operator delete[](void*, const std::nothrow_t&);             (4)   
void operator delete(void*, std::size_t)                          (5)
void operator delete[](void*, std::size_t)                        (6)
void operator delete(void*, std::size_t, const std::nothrow_t&)   (7)
void operator delete[](void*, std::size_t, const std::nothrow_t&) (8)

(5-8) Called instead of (1-4) if a user-defined replacement is provided. The standard library implementations are identical to (1-4).

I believe the cited statement is correct; however, I cannot confirm it as per the draft C++14 standard n3797 .

I examined 3.7.4 and 18.6, and found nothing to explicitly require the delete expression must call void ::operator delete(void*, std::size_t) instead of void ::operator delete(void*) if the former exists.

Could you refer me to the right page of the draft standard?

Answer 1

The selection of deallocation functions is explained in:

5.3.5 Delete [expr.delete]

10 If the type is complete and if deallocation function lookup finds both a usual deallocation function with only a pointer parameter and a usual deallocation function with both a pointer parameter and a size parameter, then the selected deallocation function shall be the one with two parameters. Otherwise, the selected deallocation function shall be the function with one parameter.