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

.

<code>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);
}
</code>

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

.

<code>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);
}
</code>

[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.

<code>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);
}
</code>

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.

<code>i = i - 1;
</code>

4. Use Reverse for Loop

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

<code>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);
}
</code>

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