c++vectoriteratorlanguage-lawyererase

Erasing vector::end from vector

Does it works correctly (does nothing) when I use

 vector<T> v;
 v.erase(v.end());

I want to use something like

 v.erase(std::find(...));

Should I if is it v.end() or not?
There is no info about it on C++.com and CPPreference

Solution

  • The standard doesn't quite spell it out, but v.erase(q) is defined, "Erases the element pointed to by q" in [sequence.reqmts]. This means that q must actually point to an element, which the end iterator doesn't. Passing in the end iterator is undefined behavior.

    Unfortunately, you need to write:

    auto it = std::find(...);
if (it != <the part of ... that specifies the end of the range searched>) {
    v.erase(it);
}

    Of course, you could define:

    template typename<Sequence, Iterator>
Iterator my_erase(Sequence &s, Iterator it) {
    if (it == s.end()) return it;
    return s.erase(it);
}

my_erase(v, std::find(v.begin(), v.end(), whatever));

    c.erase() on an associative container returns void, so to generalize this template to all containers you need some -> decltype action.