Re: an iterator question

From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 15 Jun 2007 09:11:17 -0400

Message-ID:

<f4u35n$f2r$1@news.datemas.de>

Jess wrote:

Hello,

I learned that if we do "v.end()", then the returned iterator is a
temporary object and hence cannot be changed like

--v.end();

Why? It can. If an object is *temporary* it does NOT mean that you
cannot change it. The problem here is that the value is effectively
lost right after you change it, unless you store it somewhere, like

    it = --v.end();

Try this program in "debug" mode:
----------
    #include <list>
    #include <cassert>

    int main()
    {
        std::list<int> li;
        li.push_back(1);
        li.push_back(2);
        std::list<int>::iterator it = --li.end();
        assert(*it == 2);
    }
----------
Does the assertion fail?

Why is the returned iterator a temporary pointer?

A temporary pointer? WHAT???

I think when the
result of a function is returned, then a non-temporary object is
created, with value copied from the local temporary object. If the
code above still gets us a temporary object, then I think the
non-temp- object-creation hasn't taken place yet. Does it only
happen if we do

p = v.end();

ie. assign v.end() to some variable?

Huh? You lost me.

It is also said that we can't modify it because it has a built in
type.

"It is also said" *where*?? What is it that you're reading that is
so wrong?

I guess "build-in" means pointer type here. This sounds to me
as if we could modify the temporary object if it is of class type.
I'm quite confused....

It makes two of us.

V
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