Question

Suppose I have the following code:

void* my_alloc (size_t size)
{
   return new char [size];
}

void my_free (void* ptr)
{
   delete [] ptr;
}

Is this safe? Or must ptr be cast to char* prior to deletion?

Answer 1

It depends on “safe.” It will usually work because information is stored along with the pointer about the allocation itself, so the deallocator can return it to the right place. In this sense it is “safe” as long as your allocator uses internal boundary tags. (Many do.)

However, as mentioned in other answers, deleting a void pointer will not call destructors, which can be a problem. In that sense, it is not “safe.”

There is no good reason to do what you are doing the way you are doing it. If you want to write your own deallocation functions, you can use function templates to generate functions with the correct type. A good reason to do that is to generate pool allocators, which can be extremely efficient for specific types.

As mentioned in other answers, this is undefined behavior in C++. In general it is good to avoid undefined behavior, although the topic itself is complex and filled with conflicting opinions.