Can you remove elements from a std::list while iterating through it?

Question

I\'ve got code that looks like this:

for (std::list::iterator i=items.begin();i!=items.end();i++)
{
    bool isActive = (*i)->update();
    /         


        

Answer 1

You can write

std::list::iterator i = items.begin();
while (i != items.end())
{
    bool isActive = (*i)->update();
    if (!isActive) {
        i = items.erase(i); 
    } else {
        other_code_involving(*i);
        i++;
    }
}

You can write equivalent code with std::list::remove_if, which is less verbose and more explicit

items.remove_if([] (item*i) {
    bool isActive = (*i)->update();
    if (!isActive) 
        return true;

    other_code_involving(*i);
    return false;
});

The std::vector::erase std::remove_if idiom should be used when items is a vector instead of a list to keep compexity at O(n) - or in case you write generic code and items might be a container with no effective way to erase single items (like a vector)

items.erase(std::remove_if(begin(items), end(items), [] (item*i) {
    bool isActive = (*i)->update();
    if (!isActive) 
        return true;

    other_code_involving(*i);
    return false;
}));