Why Using forEach to Remove Elements from an ArrayList Throws ConcurrentModificationException and How to Delete Safely

Programmers with 1‑3 years of experience are considered junior, while those with more experience are senior, but a recent code review revealed a five‑year‑veteran using ArrayList with forEach to delete elements, which leads to a runtime error.

1. Reproduction

The following code (with sensitive parts anonymized) demonstrates the issue:

public static void main(String[] args) {
    ArrayList
list = new ArrayList<>(Arrays.asList("1", "2", "3"));
    list.forEach(item -> {
        if (item.startsWith("1")) {
            list.remove(item);
        }
    });
}

Running this code throws a ConcurrentModificationException because the list is modified while it is being iterated.

2. Cause Analysis

The forEach construct is syntactic sugar; after compilation it becomes equivalent to the following loop:

// This is the sugar expanded by the compiler
for (Iterator i = lists.iterator(); i.hasNext(); ) {
    String s = (String) i.next();
    if (s.startsWith("1")) {
        list.remove(s);
    }
}

The Iterator.next() method checks for concurrent modification and throws ConcurrentModificationException if the underlying collection has been altered, because ArrayList 's iterator is designed to be fail‑fast.

3. Correct Deletion Methods

3.1 Use Iterator.remove

Iterator
iterator = list.iterator();
while (iterator.hasNext()) {
    String item = iterator.next();
    if (item.startsWith("1")) {
        iterator.remove();
    }
}

This approach removes elements without breaking the iterator's state.

3.2 Use removeIf (Java 8+)

list.removeIf(item -> item.startsWith("1"));

removeIf provides a concise, safe way to delete matching elements.

3.3 Collect then removeAll

List
itemsToRemove = list.stream()
    .filter(item -> item.startsWith("1"))
    .collect(Collectors.toList());
list.removeAll(itemsToRemove);

This method first gathers the elements to be removed and then deletes them in a single operation.