Re: can a const pointer be deleted?

From:
abir <abirbasak@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 2 Dec 2008 21:01:07 -0800 (PST)
Message-ID:
<44c46ad4-d318-4ac4-9a4f-0e15a62865a7@x16g2000prn.googlegroups.com>
On Dec 2, 8:05 pm, Maxim Yegorushkin <maxim.yegorush...@gmail.com>
wrote:

On Dec 2, 1:55 pm, abir <abirba...@gmail.com> wrote:

Hi,
  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.


Though the MSDN document (I am using VC9 express) says it it an
iterator
http://msdn.microsoft.com/en-us/library/ceh559x2(VS.80).aspx
and SGI is saying the same, the code says it is an const iterator and
thus,
 std::vector<int> v;
 v.push_back(1);
 const std::vector<int>& cv = v;
 std::cout<<typeid(cv.begin()).name()<<"\n";
 v.erase(cv.begin());
This code runs perfectly, and indeed cv.begin() is const_iterator and
i think
it is not possible to convert a const_iterator to iterator, though the
other way is possible.
So not sure if i am missing something or it is bug from msvc side.
Will check it on gcc.

  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 &
deallocate).
 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.


That is what i do. But my point is, if const pointer can be destroyed
(though can't be modified)
so why can't const_iterator do the same?

 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);
    std::free(ch);
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/
delete-expression:

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


I understand that. What i want to know whether delete expression after
calling destructor
implicitly cast the const pointer to non const before calling delete
operator? as delete operator
takes a void* (and not a const void*) and the pointer available after
destruction is const T* (T is the object type) which can be converted
to const void* but not to void* in 'ordinary cases'.
It seems delete expression, as you said does that extra thing to
remove const-ness and volatile
from the pointer before calling delete operator (Though i don't know
why!).

Thanks for reply.
abir

--
Max

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