C++ std::set::erase 与 std::remove_if

Question

This code has the Visual Studio error C3892 . If I change std::set to std::vector - it works.

std::set<int> a;
a.erase(std::remove_if(a.begin(), a.end(), [](int item)
{
    return item == 10;
}), a.end());

What's wrong? Why can't I use std::remove_if with std::set ?

Answer 1

You cannot use std::remove_if() with sequences which have const parts. The sequence of std::set<T> elements are made up of T const objects. We actually discussed this question just yesterday at the standard C++ committee and there is some support to create algorithms dealing specifically with the erase() ing objects from containers. It would look something like this (see also N4009 ):

template <class T, class Comp, class Alloc, class Predicate>
void discard_if(std::set<T, Comp, Alloc>& c, Predicate pred) {
    for (auto it{c.begin()}, end{c.end()}; it != end; ) {
        if (pred(*it)) {
            it = c.erase(it);
        }
        else {
            ++it;
        }
    }
}

(it would probably actually delegate to an algorithm dispatching to the logic above as the same logic is the same for other node-based container).

For you specific use, you can use

a.erase(10);

but this only works if you want to remove a key while the algorithm above works with arbitrary predicates. On the other hand, a.erase(10) can take advantage of std::set<int> 's structure and will be O(log N) while the algorithm is O(N) (with N == s.size() ).

Answer 2

std::remove_if re-orders elements, so it cannot be used with std::set . But you can use std::set::erase :

std::set<int> a;
a.erase(10);

Answer 3

Starting with C++20, you can use std::erase_if for containers with an erase() method, just as Kühl explained. Notice that this also includes std::vector , as it has an erase method. No more chaining a.erase(std::remove_if(... :)