new[n] and delete every location with delete instead the whole chunk with delete[]

Question

Is this valid C++ (e.g. not invoking UB) and does it achieve what I want without leaking memory? valgrinds complains about mismatching free and delete but says "no leaks are possible" in the end.

int main() {
  int* a = new int[5];
  for(int i = 0; i < 5; ++i)
    a[i] = i;

  for(int i = 0; i < 5; ++i)
    delete &a[i];
}

The reason I'm asking: I have a class that uses boost::intrusive::list and I new every object that is added to that list. Sometimes I know how many objects I want to add to the list and was thinking about using new[] to allocate a chunk and still be able to delete every object on its own with the Disposer-style of boost::intrusive.

Answer 1

No way. You cannot call delete on what was not allocated by new or you get heap corruption.

You see, that array created by new[] didn't allocate n individual objects, but one array. The second object of the array is in the middle of the allocation block.

Answer 2

You can add a bit so each object remembers whether it was the first in the contiguous array. This is tricky:

• totally incompatible with delete object syntax
• incompatible with inheritance
• disposer instead calls destructor explicitly, eg object.~ListItem()
• must still use array operator new[]
• be sure to dispose objects in reverse order, unlike your included example
• destructor implements reflective ownership of the array by its first element:

.

ListItem::~ListItem() {
    if ( m_own_my_subarray ) {
        operator delete[]( this ); // does not invoke any destructor!
                 // "this" must be exactly the result of new[]!
    }
}