Does C++14 require that the delete expression must call `void operator ::delete(void*, std::size_t)` instead of `void ::operator delete(void*)`?
c++, c++14, memory-management, new-operator
Solution
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.
Problem
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?