Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Deleting elements from std::set while iterating

People also ask

How do you remove from set while iterating?

Using an iterator We can use the remove() method provided by the Iterator interface that removes the latest element returned by the iterator. Please note we should not modify the set after the iterator is created (except through the iterator's own remove method); otherwise, a ConcurrentModificationException is thrown.

How do you remove an element from a set in C++?

set::erase() erase() function is used to remove elements from a container from the specified position or range.

How do you remove an element from a list in a while loop?

If you want to delete elements from a list while iterating, use a while-loop so you can alter the current index and end index after each deletion.

Can we erase element from set?

Sets are a type of associative container in which each element has to be unique because the value of the element identifies it. The value of the element cannot be modified once it is added to the set, though it is possible to remove and add the modified value of that element.


This is implementation dependent:

Standard 23.1.2.8:

The insert members shall not affect the validity of iterators and references to the container, and the erase members shall invalidate only iterators and references to the erased elements.

Maybe you could try this -- this is standard conforming:

for (auto it = numbers.begin(); it != numbers.end(); ) {
    if (*it % 2 == 0) {
        numbers.erase(it++);
    }
    else {
        ++it;
    }
}

Note that it++ is postfix, hence it passes the old position to erase, but first jumps to a newer one due to the operator.

2015.10.27 update: C++11 has resolved the defect. iterator erase (const_iterator position); return an iterator to the element that follows the last element removed (or set::end, if the last element was removed). So C++11 style is:

for (auto it = numbers.begin(); it != numbers.end(); ) {
    if (*it % 2 == 0) {
        it = numbers.erase(it);
    }
    else {
        ++it;
    }
}

If you run your program through valgrind, you'll see a bunch of read errors. In other words, yes, the iterators are being invalidated, but you're getting lucky in your example (or really unlucky, as you're not seeing the negative effects of undefined behavior). One solution to this is to create a temporary iterator, increment the temp, delete the target iterator, then set the target to the temp. For example, re-write your loop as follows:

std::set<int>::iterator it = numbers.begin();                               
std::set<int>::iterator tmp;                                                

// iterate through the set and erase all even numbers                       
for ( ; it != numbers.end(); )                                              
{                                                                           
    int n = *it;                                                            
    if (n % 2 == 0)                                                         
    {                                                                       
        tmp = it;                                                           
        ++tmp;                                                              
        numbers.erase(it);                                                  
        it = tmp;                                                           
    }                                                                       
    else                                                                    
    {                                                                       
        ++it;                                                               
    }                                                                       
} 

You misunderstand what "undefined behavior" means. Undefined behavior does not mean "if you do this, your program will crash or produce unexpected results." It means "if you do this, your program could crash or produce unexpected results", or do anything else, depending on your compiler, your operating system, the phase of the moon, etc.

If something executes without crashing and behaves as you expect it to, that is not proof that it is not undefined behavior. All it proves is that its behavior happened to be as observed for that particular run after compiling with that particular compiler on that particular operating system.

Erasing an element from a set invalidates the iterator to the erased element. Using an invalidated iterator is undefined behavior. It just so happened that the observed behavior was what you intended in this particular instance; it does not mean that the code is correct.


C++20 will have "uniform container erasure", and you'll be able to write:

std::erase_if(numbers, [](int n){ return n % 2 == 0 });

And that will work for vector, set, deque, etc. See cppReference for more info.


Just to warn, that in case of a deque container, all solutions that check for the deque iterator equality to numbers.end() will likely fail on gcc 4.8.4. Namely, erasing an element of the deque generally invalidates pointer to numbers.end():

#include <iostream>
#include <deque>

using namespace std;
int main() 
{

  deque<int> numbers;

  numbers.push_back(0);
  numbers.push_back(1);
  numbers.push_back(2);
  numbers.push_back(3);
  //numbers.push_back(4);

  deque<int>::iterator  it_end = numbers.end();

  for (deque<int>::iterator it = numbers.begin(); it != numbers.end(); ) {
    if (*it % 2 == 0) {
      cout << "Erasing element: " << *it << "\n";
      numbers.erase(it++);
      if (it_end == numbers.end()) {
    cout << "it_end is still pointing to numbers.end()\n";
      } else {
    cout << "it_end is not anymore pointing to numbers.end()\n";
      }
    }
    else {
      cout << "Skipping element: " << *it << "\n";
      ++it;
    }
  }
}

Output:

Erasing element: 0
it_end is still pointing to numbers.end()
Skipping element: 1
Erasing element: 2
it_end is not anymore pointing to numbers.end()

Note that while the deque transformation is correct in this particular case, the end pointer has been invalidated along the way. With the deque of a different size the error is more apparent:

int main() 
{

  deque<int> numbers;

  numbers.push_back(0);
  numbers.push_back(1);
  numbers.push_back(2);
  numbers.push_back(3);
  numbers.push_back(4);

  deque<int>::iterator  it_end = numbers.end();

  for (deque<int>::iterator it = numbers.begin(); it != numbers.end(); ) {
    if (*it % 2 == 0) {
      cout << "Erasing element: " << *it << "\n";
      numbers.erase(it++);
      if (it_end == numbers.end()) {
    cout << "it_end is still pointing to numbers.end()\n";
      } else {
    cout << "it_end is not anymore pointing to numbers.end()\n";
      }
    }
    else {
      cout << "Skipping element: " << *it << "\n";
      ++it;
    }
  }
}

Output:

Erasing element: 0
it_end is still pointing to numbers.end()
Skipping element: 1
Erasing element: 2
it_end is still pointing to numbers.end()
Skipping element: 3
Erasing element: 4
it_end is not anymore pointing to numbers.end()
Erasing element: 0
it_end is not anymore pointing to numbers.end()
Erasing element: 0
it_end is not anymore pointing to numbers.end()
...
Segmentation fault (core dumped)

Here is one of the ways to fix this:

#include <iostream>
#include <deque>

using namespace std;
int main() 
{

  deque<int> numbers;
  bool done_iterating = false;

  numbers.push_back(0);
  numbers.push_back(1);
  numbers.push_back(2);
  numbers.push_back(3);
  numbers.push_back(4);

  if (!numbers.empty()) {
    deque<int>::iterator it = numbers.begin();
    while (!done_iterating) {
      if (it + 1 == numbers.end()) {
    done_iterating = true;
      } 
      if (*it % 2 == 0) {
    cout << "Erasing element: " << *it << "\n";
      numbers.erase(it++);
      }
      else {
    cout << "Skipping element: " << *it << "\n";
    ++it;
      }
    }
  }
}

This behaviour is implementation specific. To guarantee the correctness of the iterator you should use "it = numbers.erase(it);" statement if you need to delete the element and simply incerement iterator in other case.