Allocate n bytes by new[] and fill it with any type?

Question

I want to dynamically allocate known size of memory (just memory, not bothering about type) and fill it with exactly the same amount of data but any type (I'm only sure it will be primitive type). Ofc later I'm going to free it.

Is it ok? :

auto dest = new int8_t[n];
std::memcpy(dest, src, n);
delete[] dest;

src is ptr to an array of size n (Bytes). I've ofc chosen int8_t becuase it's the clearest way to allocate certain amount of memory. In fact the code above isn't exaclt what it will be. delete[] will be called on pointer of type which actually it points to. So for example if src was an array of floats (forget about last line of above code):

float * ptr = dest;
delete[] ptr;

So again. Will it be ok?

Answer 1

It is ok, but only if you use a char array or an unsigned char array, because of the special alignment guarantees for these types:

5.3.4 New
11 When a new-expression calls an allocation function and that allocation has not been extended, the new- expression passes the amount of space requested to the allocation function as the first argument of type std::size_t. That argument shall be no less than the size of the object being created; it may be greater than the size of the object being created only if the object is an array. For arrays of char and unsigned char, the difference between the result of the new-expression and the address returned by the allocation function shall be an integral multiple of the strictest fundamental alignment requirement (3.11) of any object type whose size is no greater than the size of the array being created. [ Note: Because allocation functions are assumed to return pointers to storage that is appropriately aligned for objects of any type with fundamental alignment, this constraint on array allocation overhead permits the common idiom of allocating character arrays into which objects of other types will later be placed. — end note ]

emphasis by me.

Another requirement is that you only use primitive types or PODs, because you don't call a constructor, but the (trivial) destructor (through delete).