Why [] is used in delete ( delete [] ) to free dynamically allocated array ?

Question

I know that when delete [] will cause destruction for all array elements and then releases the memory.

I initially thought that compiler wants it just

Answer 1

The heap itself knows the size of allocated block - you only need the address. Look like free() works - you only pass the address and it frees memory.

The difference between delete (delete[]) and free() is that the former two first call the destructors, then free memory (possibly using free()). The problem is that delete[] also has only one argument - the address and having only that address it need to know the number of objects to run destructors on. So new[] uses som implementation-defined way of writing somewhere the number of elements - usually it prepends the array with the number of elements. Now delete[] will rely on that implementation-specific data to run destructors and then free memory (again, only using the block address).