Avoid ConcurrentModificationException: Safe Ways to Remove Elements While Iterating in Java

1. Common Mistake by Beginners

Many beginners try to remove elements from a List inside a foreach loop, which immediately throws java.util.ConcurrentModificationException.

public static void main(String[] args) {
    List<String> platformList = new ArrayList<>();
    platformList.add("博客园");
    platformList.add("CSDN");
    platformList.add("掘金");

    for (String platform : platformList) {
        if (platform.equals("博客园")) {
            platformList.remove(platform);
        }
    }

    System.out.println(platformList);
}

The foreach construct is compiled to use an Iterator whose core methods are hasNext() and next(). During each next() call the iterator invokes checkForComodification(), which compares the collection’s internal modCount with an expectedModCount. Removing an element directly from the list changes modCount but not expectedModCount, causing the mismatch and the exception.

Therefore, foreach cannot be used for removal.

2. Use Iterator.remove() Method

Iterating with an explicit Iterator and calling its remove() safely updates expectedModCount.

public static void main(String[] args) {
    List<String> platformList = new ArrayList<>();
    platformList.add("博客园");
    platformList.add("CSDN");
    platformList.add("掘金");

    Iterator<String> iterator = platformList.iterator();
    while (iterator.hasNext()) {
        String platform = iterator.next();
        if (platform.equals("博客园")) {
            iterator.remove();
        }
    }

    System.out.println(platformList);
}

[CSDN, 掘金]

Each call to iterator.remove() resets expectedModCount to the current modCount, keeping them equal and preventing the exception.

3. Use Forward for Loop

Iterating with an index allows removal, but the index must be adjusted after each deletion.

public static void main(String[] args) {
    List<String> platformList = new ArrayList<>();
    platformList.add("博客园");
    platformList.add("CSDN");
    platformList.add("掘金");

    for (int i = 0; i < platformList.size(); i++) {
        String item = platformList.get(i);
        if (item.equals("博客园")) {
            platformList.remove(i);
            i = i - 1;
        }
    }

    System.out.println(platformList);
}

After removing an element the list shrinks, so the next element shifts to the current index; decrementing i ensures the shifted element is not skipped.

i = i - 1;

4. Use Reverse for Loop

Iterating from the end avoids index correction because removal does not affect the yet‑to‑visit elements.

public static void main(String[] args) {
    List<String> platformList = new ArrayList<>();
    platformList.add("博客园");
    platformList.add("CSDN");
    platformList.add("掘金");

    for (int i = platformList.size() - 1; i >= 0; i--) {
        String item = platformList.get(i);
        if (item.equals("掘金")) {
            platformList.remove(i);
        }
    }

    System.out.println(platformList);
}

Since the loop proceeds backward, removing an element does not shift the indices of the remaining elements that are still to be processed.