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How to do it...

In this section, we are filling an std::vector instance with some example integers, and then prune some specific items away from it. The way we are doing it is considered the correct way of removing multiple items from a vector.

Of course we need to include some headers before we do anything.

      #include <iostream>
      #include <vector>
      #include <algorithm>

Then we declare that we are using namespace std to spare us some typing.

      using namespace std;

Now we create us a vector of integers and fill it with some example items.

      int main()
      {
          vector<int> v {1, 2, 3, 2, 5, 2, 6, 2, 4, 8};

The next step is to remove the items. What do we remove? There are multiple 2 values. Let's remove them.

          const auto new_end (remove(begin(v), end(v), 2));

Interestingly, that was only one of the two steps. The vector still has the same size. The next line makes it actually shorter.

          v.erase(new_end, end(v));

Let's stop by here in order to print the vector's content to the terminal, and then continue.

          for (auto i : v) {
              cout << i << ", ";
          }
          cout << '\n';

Now, let's remove a whole class of items, instead of specific values. In order to do that, we define a predicate function first, which accepts a number as parameter, and returns true, if it is an odd number.

          const auto odd ([](int i) { return i % 2 != 0; });

Now we use the remove_if function and feed it with the predicate function. Instead of removing in two steps as we did before, we do it in one.

          v.erase(remove_if(begin(v), end(v), odd), end(v));

All odd items are gone now, but the vector's capacity is still at the old 10 elements. In a last step, we reduce that also to the actual current size of the vector. Note that this might lead the vector code to allocate a new chunk of memory that fits and moves all items from the old chunk to the new one.

          v.shrink_to_fit();

Now, let's print the content after the second run of removing items and that's it.

          for (auto i : v) {
              cout << i << ", ";
          }
          cout << '\n';
      }

Compiling and running the program yields the following two output lines from the two item removing approaches.

      $ ./main 
      1, 3, 5, 6, 4, 8, 
      6, 4, 8,