Why does my vector not re-size properly after erasing items?

Question

I wrote a small program to remove 0's from a vector. But when I check, there is still a zero in the vector. why?

The size says its 4 when it should be 3 and there is still a zero showing.

int main()
{

    vector<int> nums1 = {1, 2, 3, 0, 0, 0};

    for(int i=0; i< nums1.size(); i++)
    {
        if(nums1[i] == 0)
            nums1.erase(nums1.begin() + i);
    }
    cout << "size is now: " << nums1.size() << endl;

    for(int j=0; j<nums1.size(); j++)
        cout<< nums1[j] << endl;
    return 0;
}

You shouldn't increment `i` when the iteration erased an element. Otherwise you skip the element that now occupies the position of the element you removed. You should step through your code with a debugger to see what happens. — François Andrieux, Aug 20 '21 at 20:55
Also I'd recommend to read about [iterators](https://www.geeksforgeeks.org/introduction-iterators-c/). — Karen Baghdasaryan, Aug 20 '21 at 21:00
Use an iterator for enumeration and take advantage of the return value of `erase`. Better still, use the [remove/erase idiom](https://stackoverflow.com/questions/347441/erasing-elements-from-a-vector) — WhozCraig, Aug 20 '21 at 21:01
You may want to read [How do I remove an item from a stl vector with a certain value?](https://stackoverflow.com/questions/39912/how-do-i-remove-an-item-from-a-stl-vector-with-a-certain-value) for a better way to do this. — Retired Ninja, Aug 20 '21 at 21:01
If you’re going to be removing items from the vector you might consider iterating backwards (eg for (int i = v.size()-1; i >= 0; i—) {…} ), that way you don’t have make your integer-index modification conditional. — Jeremy Friesner, Aug 20 '21 at 21:12
"_to remove 0's from a vector_" - Since C++20: [`std::erase/erase_if(std::vector)`](https://en.cppreference.com/w/cpp/container/vector/erase2) — Ted Lyngmo, Aug 20 '21 at 21:14

score 5 · Answer 1 · answered Aug 20 '21 at 21:15

5

As mentioned in the comment, when you erase, your vector is modified, so you must not increment your iterator in the same way.

A more idiomatic way is:

for(auto i=nums1.begin(); i != nums1.end(); )
{
    if(*i == 0)
        i = nums1.erase(i); // when erased, move to element after erased one
    else
        i++; // otherwise, increment
}

answered Aug 20 '21 at 21:15

Jeffrey

11,063
1
21
42

Ty for the explanation. I was incrementing incorrectly. Much appreciated. – user14377467 Aug 21 '21 at 16:09

score 5 · Accepted Answer · answered Aug 20 '21 at 21:47

When an element of the vector is erased then all elements that follow it are moved left occupying the position of the erased element.

That is if the element at the position i was erased then the element that follows it now will be at this position. So you shall not increase the position.

For example

for(  std::vector<int>::size_type i=0; i < nums1.size(); )
{
    if(nums1[i] == 0)
        nums1.erase(nums1.begin() + i);
    else
        ++i;
}

Pay attention to that before the C++ 20 Standard you could use the standard algorithm std::remove like

nums1.erase( std::remove( nums1.begin(), nums1.end(), 0 ), nums1.end() );

The second function you wrote is awesome. I am new to using vectors so thank you for the explaination. — user14377467, Aug 21 '21 at 16:10

score 2 · Answer 3 · answered Aug 20 '21 at 21:34

2

It's also worth mentioning, if you're using C++20 the standard library has improved support for this, which is much more clean, succinct, and less error prone:

#include <vector>
#include <algorithm>

int main() {
    std::vector v{1, 2, 0, 0, 0, 5};
    std::erase(v, 0);
}

Leaving v holding: {1, 2, 5}

answered Aug 20 '21 at 21:34

Chris Uzdavinis

6,022
9
16

Thank you for this. I am currently compiling with c++11. I will update my ide to compile with c++20. – user14377467 Aug 21 '21 at 16:13

Why does my vector not re-size properly after erasing items?

3 Answers3