Removing item from vector while iterating?

Question

I have a vector that holds items that are either active or inactive. I want the size of this vector to stay small for performance issues, so I want items that have been marked inactive to be erased from the vector. I tried doing this while iterating but I am getting the error "vector iterators incompatible".

vector<Orb>::iterator i = orbsList.begin();
    while(i != orbsList.end()) {
        bool isActive = (*i).active;
        if(!isActive) {
            orbsList.erase(i++);
        }
        else {
            // do something with *i
            ++i;
        }
    }

Solution 1

The most readable way I've done this in the past is to use std::vector::erase combined with std::remove_if. In the example below, I use this combination to remove any number less than 10 from a vector.

(For non-c++0x, you can just replace the lambda below with your own predicate:)

// a list of ints
int myInts[] = {1, 7, 8, 4, 5, 10, 15, 22, 50. 29};
std::vector v(myInts, myInts + sizeof(myInts) / sizeof(int));
// get rid of anything < 10
v.erase(std::remove_if(v.begin(), v.end(), 
                       [](int i) { return i < 10; }), v.end());

Solution 2

I agree with wilx's answer. Here is an implementation:

// curFiles is: vector < string > curFiles;
vector< string >::iterator it = curFiles.begin();
while(it != curFiles.end()) {
    if(aConditionIsMet) {
        it = curFiles.erase(it);
    }
    else ++it;
}

Solution 3

You can do that but you will have to reshuffle your while() a bit, I think. The erase() function returns an iterator to the element next after the erased one: iterator erase(iterator position);. Quoting from the standard from 23.1.1/7:

The iterator returned from a.erase(q) points to the element immediately following q prior to the element being erased. If no such element exists, a.end() is returned.

Though maybe you should be using the Erase-remove idiom instead.

Solution 4

erase returns a pointer to the next iterator value (same as Vassilis):

vector <cMyClass>::iterator mit
for(mit = myVec.begin(); mit != myVec.end(); )
{   if(condition)
        mit = myVec.erase(mit);
    else
        mit++;
}

Solution 5

If someone need working on indexes

vector<int> vector;
for(int i=0;i<10;++i)vector.push_back(i);
int size = vector.size();
for (int i = 0; i < size; ++i)
{
    assert(i > -1 && i < (int)vector.size());
    if(vector[i] % 3 == 0)
    {
        printf("Removing %d, %d\n",vector[i],i);
        vector.erase(vector.begin() + i);
    }
    if (size != (int)vector.size())
    {
        --i;
        size = vector.size();
        printf("Go back %d\n",size);
    }
}

View more solutions

Comments

[harper]

Using the return value of erase you get a valid iterator back. So it's only necessary to assign the returned value to the iterator -- no performance issue here.

[Kiran Kumar]

Also, to downsize the vector(as it seems to what the OP wants to do) the swap trick can be used. Details here:stackoverflow.com/questions/253157/how-to-downsize-stdv‌​ector

[Oliver Charlesworth]

@harper: Actually, there's a big performance issue here; erasing an item from the middle of a vector requires all the rest of them to be moved down, which involves O(N) constructor and destructor calls each time.

[harper]

And the use of an index has what befinit? How is the erasing performance increased? I just annotated the statement "iterators get invalidated .. no matter which form"...

[jakebman]

It might be enough to say i=orbsList.erase(i) instead of orbsList.erase(i++)

[EFreak]

I know its a little late but you might wanna read this. Its contrary to what most people would expect but it comes from the creator of C++ bulldozer00.com/2012/02/09/vectors-and-lists

[Josh Sanders]

While elegant in the simple case, I'm dubious of its effectiveness in the real world. This wouldn't even work at all with a dynamic size vector.

[user1122069]

Why not just size-- i-- next to erase?

[jhasse]

"This wouldn't even work at all with a dynamic size vector." Why not?

[Gelldur]

@user1122069 my good practices forbids such things, a lot of people do not understand difference between --i nad i--

[truthadjustr]

While the std::remove_if lambda looks nice, but if I only want to delete exactly one element that matches the condition, it still goes through each element to the end. Therefore, I prefer manual iteration so I can break from the loop any time.

[Eric Reed]

access the value at the iterator position using *mit