C++ Erase vector element by value rather than by position?

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Answers


How about std::remove() instead:

#include <algorithm>
...
vec.erase(std::remove(vec.begin(), vec.end(), 8), vec.end());

This combination is also known as the erase-remove idiom.


You can use std::find to get an iterator to a value:

#include <algorithm>
std::vector<int>::iterator position = std::find(myVector.begin(), myVector.end(), 8);
if (position != myVector.end()) // == myVector.end() means the element was not found
    myVector.erase(position);

You can not do that directly. You need to use std::remove algorithm to move the element to be erased to the end of the vector and then use erase function. Something like: myVector.erase(std::remove(myVector.begin(), myVector.end(), 8), myVec.end());. See this erasing elements from vector for more details.


Eric Niebler is working on a range-proposal and some of the examples show how to remove certain elements. Removing 8. Does create a new vector.

#include <iostream>
#include <range/v3/all.hpp>

int main(int argc, char const *argv[])
{
    std::vector<int> vi{2,4,6,8,10};
    for (auto& i : vi) {
        std::cout << i << std::endl;
    }
    std::cout << "-----" << std::endl;
    std::vector<int> vim = vi | ranges::view::remove_if([](int i){return i == 8;});
    for (auto& i : vim) {
        std::cout << i << std::endl;
    }
    return 0;
}

outputs

2 4 6 8 10 ----- 2 4 6 10


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