Re: an iterator question
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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