Question

I saw an example of using the function: delete in cpp and I didn’t completely understand it.
the code is:

class Name {
    const char* s;
    //...
};

class Table {
      Name* p;
      size_t sz;
public:
      Table(size_t s = 15){p = new Name[sz = s]; }
      ~Table { delete[] p; }
};

What is the exact action of the command: delete[] p;?

I think the aim was to delete all the pointers in the container Table.

The brackets in delete[] give me a clue that it deletes an array of pointers to Name but the size of the array is not specified, so how does the destructor “know” how many pointers to delete?

Answer 1

delete isn’t a function, it’s an operator.

A delete expression using [] destroys objects created with new ... [] and releases the associated memory. delete[] must be used for pointers returned by new ... []; non-array delete only on pointers returned by non-array new. Using the non-matching delete form is always incorrect.

The delete expression in ~Table() (missing () in your code) will destroy the dynamically created array of Name objects ensuring that the Name destructor is called for each member of the array.

It is up the the implementation to implement some mechanism of recording the number of elements in arrays allocated with new ... [] the programmer doesn’t have to worry about this.

In many implementations, where the array elements have non-trivial destructors, a new[] expression will allocate extra space to record the element count before the space for all the array members. This hidden count is then looked up when delete[] is used to ensure the correct number of destructors are called. This is just an implementation detail, though, other implementations are possible.