I want to replace one element in a list<string> by a list<string> and
I need to obtain an iterator to the first element in the inserted
list.
In code:

void replace_element_by_list(list<string> &my_list,
list<string>::iterator
&iter_to_remove,
const list<string>
&list_to_insert) {
list<string>::iterator next_iter = my_list.erase(iter_to_remove);
my_list.insert(next_iter, list_to_insert.begin(),
list_to_insert.end());
// Does 'iter_to_remove' point to the first element of the inserted
list, or is it invalid?
}

When I try this code I find that 'iter_to_remove' indeed points to the
first element of the inserted list. However, I'm wondering whether
that is guaranteed.

JurgenvonOerthel@hotmail.com wrote:
> I want to replace one element in a list<string> by a list<string> and
> I need to obtain an iterator to the first element in the inserted
> list.
> In code:
>
> void replace_element_by_list(list<string> &my_list,
> list<string>::iterator
> &iter_to_remove,
> const list<string>
> &list_to_insert) {
> list<string>::iterator next_iter = my_list.erase(iter_to_remove);

Add
list<string>::iterator prev_iter = next_iter;
if (prev_iter == my_list.begin())
prev_iter = my_list.end();
else
--prev_iter;

> my_list.insert(next_iter, list_to_insert.begin(),
> list_to_insert.end());

What does 'insert' return?

> // Does 'iter_to_remove' point to the first element of the inserted
> list, or is it invalid?

'iter_to_remove' is invalid. However, you can return 'prev_iter' here.
The catch, of course, is that if you're removing the very first element
in the list, you'll get 'end()'.

> }
>
> When I try this code I find that 'iter_to_remove' indeed points to the
> first element of the inserted list. However, I'm wondering whether
> that is guaranteed.

Nope.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

On Dec 7, 8:17 pm, JurgenvonOert...@hotmail.com wrote:
> I want to replace one element in a list<string> by a list<string> and
> I need to obtain an iterator to the first element in the inserted
> list.
> In code:
>
> void replace_element_by_list(list<string> &my_list,
> list<string>::iterator
> &iter_to_remove,
> const list<string>
> &list_to_insert) {
> list<string>::iterator next_iter = my_list.erase(iter_to_remove);
> my_list.insert(next_iter, list_to_insert.begin(),
> list_to_insert.end());
> // Does 'iter_to_remove' point to the first element of the inserted
> list, or is it invalid?
>
> }
>
> When I try this code I find that 'iter_to_remove' indeed points to the
> first element of the inserted list. However, I'm wondering whether
> that is guaranteed.

This is not guaranteed. It is invalid. You could, however, get the
iterator to the first element of the inserted range as below:

void replace_element_by_list(list<string> &my_list,
list<string>::iterator
&iter_to_remove,
const list<string>
&list_to_insert) {
list<string>::iterator next_iter = my_list.erase(iter_to_remove);
list<string>::iterator prev_to_next_iter = --next_iter;
my_list.insert(next_iter, list_to_insert.begin(),
list_to_insert.end());
// Does 'iter_to_remove' point to the first element of the inserted
list, or is it invalid?
iter_to_remove = ++prev_to_next_iter;
//iter_to_remove is guaranteed to point to the first element of the
//inserted list/range
}

On Dec 7, 4:30 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:
> Add
> list<string>::iterator prev_iter = next_iter;
> if (prev_iter == my_list.begin())
> prev_iter = my_list.end();
> else
> --prev_iter;
I was afraid I would have to do something ugly like this.

> What does 'insert' return?
That's my point: it returns void. I think it should return an iterator
to the first inserted element. That's what 'insert' of a single
element does.