a small issue with std::vector and changing the collection while looping through it

Question

This loop changes the iterators while running:

std::vector<int> c;
c.push_back(1);
c.push_back(2);

std::vector<int>::iterator iter    = c.begin();
std::vector<int>::iterator endIter = c.end();

while( iter != endIter )
{
    std::cout << (*iter) << std::endl;
    iter = c.erase(iter);
}

It does not work because:

Iterators and references to the erased elements and to the elements between them and the end of the container are invalidated. Past-the-end iterator is also invalidated

How can I rewrite this (without using std::list, and using the while loop) ?

By the way, I know that auto has been implemented since C++11. Why would it be beneficial to use it ?

Answer 1

Simply do not cache the end iterator that will be invalidated:

while( iter != c.end() )
{
    std::cout << (*iter) << std::endl;
    iter = c.erase(iter);
}

or clear the vector after printing:

for(const auto& i : c) {
    std::cout << i << std::endl;
}
c.clear();

Answer 2

Erasing an element changes end(). Change the loop:

while( iter != c.end())

Answer 3

Either

• Rewrite it as

  while( iter != c.end() )
  {
      std::cout << (*iter) << std::endl;
      iter = c.erase(iter);
  }
  

  and the code will no longer rely on any potentially invalidated iterators,

or

• "Refresh" any potentially invalidated iterators after each invalidating operation

  while( iter != endIter )
  {
      std::cout << (*iter) << std::endl;
      iter = c.erase(iter);
      endIter = c.end();
  }
  

These are the two generic approaches typically used in cases like that.

Answer 4

A more idiomatic way of doing this...

while(c.begin() != c.end()) c.erase(c.begin());

Though this is very slow, as a vectors underlying implementation uses a contiguous array(with extra space on the end). So repeatedly erasing the begin element is very ineficient, as every element ends up getting copied one space in the array earlier, n - index times! You can jurastically increase performance by doing this:

while(c.begin() != c.end()) c.pop_back();