Removing item from vector while iterating?

问题:

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;
        }
    }

回答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());

回答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;
}

回答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.

回答4:

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);
    }
}

回答5:

You might want to consider using a std::list instead of a std::vector for your data structure. It is safer (less bug prone) to use when combining erasure with iteration.

回答6:

As they said, vector's iterators get invalidated on vector::erase() no matter which form of iterator increment you use. Use an integer index instead.

回答7:

Removing items from the middle of a vector will invalidate all iterators to that vector, so you cannot do this (update: without resorting to Wilx's suggestion).

Also, if you're worried about performance, erasing items from the middle of a vector is a bad idea anyway. Perhaps you want to use an std::list?