Re: can a const pointer be deleted?

Maxim Yegorushkin <>
Tue, 2 Dec 2008 07:05:39 -0800 (PST)
On Dec 2, 1:55 pm, abir <> wrote:

  In the std containers erase takes a const_iterator.

This is not so.

In std containers erase member function is a non-const member
function, because it changes the container contents. It would not be
logical for that function to accept const_iterator.

  I have my own container and there const_iterator is a const pointer
to the item.
  I want to free the memory for the item on erase (i.e both destroy &
 However deallocate function for allocator (and so delete, free etc)
takes a pointer rather than
 a const pointer. Is that mean i have to cast it back to a pointer?

Make your_container::erase() accept non-const iterators.

 I was looking at std::list. There even a const_iterator returns a non
const node pointer (not very sure though).
 Also how this works ? does the operator delete automatically cast it
to a non const type ?
const testobj* to = new testobj(1,1);
delete to;

There are two different things in C++ named delete: a) delete-
expression; b) operator delete. Here you are using delete-expression
and it ignores const-ness and volatile-ness. What delete-expression a)
does is it calls the destructor of the object and then invokes
operator delete b).

In C, while this works
char* ch = (char*)std::malloc(5);
for a const pointer i explicitly need to cast it back like,
const char* ch = (const char*)std::malloc(5);
 std::free((char*)ch);//or std::free(void*)ch);

In C++ the equivalents of malloc/free are operator new/delete, not new/

    #include <new>
    int main()
        char const* c = (char const*)operator new(5);
        operator delete((void*)c); // need a cast here


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