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<item*>::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;
}));

Answer 2

Use std::remove_if algorithm.

Edit: Work with collections should be like: 1. prepare collection. 2. process collection.

Life will be easier if you won't mix this steps.

1. std::remove_if. or list::remove_if ( if you know that you work with list and not with the TCollection )
2. std::for_each

Answer 3

I think you have a bug there, I code this way:

for (std::list<CAudioChannel *>::iterator itAudioChannel = audioChannels.begin();
             itAudioChannel != audioChannels.end(); )
{
    CAudioChannel *audioChannel = *itAudioChannel;
    std::list<CAudioChannel *>::iterator itCurrentAudioChannel = itAudioChannel;
    itAudioChannel++;

    if (audioChannel->destroyMe)
    {
        audioChannels.erase(itCurrentAudioChannel);
        delete audioChannel;
        continue;
    }
    audioChannel->Mix(outBuffer, numSamples);
}

Answer 4

Iterating backwards avoids the effect of erasing an element on the remaining elements to be traversed:

typedef list<item*> list_t;
for ( list_t::iterator it = items.end() ; it != items.begin() ; ) {
    --it;
    bool remove = <determine whether to remove>
    if ( remove ) {
        items.erase( it );
    }
}

PS: see this, e.g., regarding backward iteration.

PS2: I did not thoroughly tested if it handles well erasing elements at the ends.

Answer 5

Here's an example using a for loop that iterates the list and increments or revalidates the iterator in the event of an item being removed during traversal of the list.

for(auto i = items.begin(); i != items.end();)
{
    if(bool isActive = (*i)->update())
    {
        other_code_involving(*i);
        ++i;

    }
    else
    {
        i = items.erase(i);

    }

}

items.remove_if(CheckItemNotActive);

Answer 6

If you think of the std::list like a queue, then you can dequeue and enqueue all the items that you want to keep, but only dequeue (and not enqueue) the item you want to remove. Here's an example where I want to remove 5 from a list containing the numbers 1-10...

std::list<int> myList;

int size = myList.size(); // The size needs to be saved to iterate through the whole thing

for (int i = 0; i < size; ++i)
{
    int val = myList.back()
    myList.pop_back() // dequeue
    if (val != 5)
    {
         myList.push_front(val) // enqueue if not 5
    }
}

myList will now only have numbers 1-4 and 6-10.