Oops! I was confused with English languagw again, sorry!
or a (possibly multi-dimensional) array thereof
This still refers " class type with a non-trivial destructor". Therefore,
(10.4) comes into play in my case:
(10.4) Otherwise, it is unspecified whether a deallocation function with a
parameter of type std::size_t is selected.
Thanks for help, anyway!
Yours truly,
brigadir15 at gmail dot com.
On December 4, 2016 7:50:17 PM Avi Kivity <avi@xxxxxxxxxxxx> wrote:
On 12/04/2016 02:26 PM, Ruslan Garipov wrote:
I have the following `operator delete` replacements:
void operator delete[](void* p)
{
/* Implementation does not matter. */
}
void operator delete[](void* p, std::size_t size)
{
/* Implementation does not matter. */
}
My question is why, in the following code, GCC 6.2 calls `void
operator delete[](void*)` and not the second replacement:
char* str = new char[14];
delete[] str;
According to 5.3.5 Delete [expr.delete]:
(10.3) If the type is complete and if, for the second alternative (delete
array) only, the operand is a pointer to a class type with a non-trivial
destructor or a (possibly multi-dimensional) array thereof, the function
with a parameter of type std::size_t is selected.
Therefore, I believe `operator delete[](void*, std::size_t)` must be
called, doesn't it?
char is not a class type with a non-trivial destructor, so the second
condition does not hold.
Under the hood, if C++ allocates an array with non-trivial destructors,
it must also store the size of the array so it knows how many
destructors to call, and can then use the stored size of the array to
call the sized deallocator function. In this case, there's no need to
call the destructors, so C++ only allocates 14 chars, and does not store
the size.
