gpt4 book ai didi

java - 从 ArrayList 中删除对象时如何避免并发修改异常”

转载 作者:行者123 更新时间:2023-12-02 08:50:43 25 4
gpt4 key购买 nike

考虑以下代码。

正如您所料,当在 for-each 循环内删除水果时,deleteFruitByName 方法会引发 ConcurrentModificationException。

在这种情况下我该如何避免这种情况?

import java.util.ArrayList;

public class Stringplay {
public static void main(String[] args) {
ArrayList<Fruit> fruites = new ArrayList<Fruit>();
new Fruit(32, "apple", "red");
new Fruit(64, "orange", "orange");
new Fruit(12, "banana", "red");
new Fruit(42, "grape", "purple");
fruites.addAll(Fruit.fruits);
Fruit.deleteFruitByName("apple");
for (Fruit fruit : fruites) {
System.out.println(fruit.getName());
}
}

}

public class Fruit {
public int weight;
public String name;
public String type;
public static ArrayList<Fruit> fruits = new ArrayList<Fruit>();

public Fruit(int weight, String name, String type) {
this.weight = weight;
this.name = name;
this.type = type;
fruits.add(this);
}

public String getName() {
return name;
}

public static void deleteFruitByName(String fruitName) {
for (Fruit fruit : fruits) {
if (fruit.getName().equals(fruitName)) {
fruits.remove(fruit);
}
}

}
}


最佳答案

为了避免 ConcurrentModificationException,您需要在此处使用 Iterator。

    public static void deleteFruitByName(String fruitName) {
Iterator<Fruit> it = fruits.iterator();
while (it.hasNext()) {
Fruit fruit = it.next();
if (fruit.getName().equals(fruitName)) {
it.remove();
}
}
}

来自java doc

The iterators returned by this class's iterator and listIterator methods are fail-fast: if the list is structurally modified at any time after the iterator is created, in any way except through the iterator's own remove or add methods, the iterator will throw a ConcurrentModificationException. Thus, in the face of concurrent modification, the iterator fails quickly and cleanly, rather than risking arbitrary, non-deterministic behavior at an undetermined time in the future.

更新:要迭代 Fluit 类中的集合,请使用此代码

public class Main {

public static void main(String[] args) {
new Fruit(32, "apple", "red");
new Fruit(64, "orange", "orange");
new Fruit(12, "banana", "red");
new Fruit(42, "grape", "purple");
Fruit.deleteFruitByName("apple");
for (Fruit fruit : Fruit.fruits) {
System.out.println(fruit.getName());
}
}
}

关于java - 从 ArrayList 中删除对象时如何避免并发修改异常”,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60797684/

25 4 0