我如何克隆一个数组列表,也克隆其项目在Java?

例如,我有:

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = ....something to do with dogs....

我希望clonedList中的对象与dogs list中的对象不一样。


当前回答

我认为目前的绿色答案很糟糕,为什么你会问?

它可能需要添加大量代码 它要求你列出所有要复制的列表并这样做

序列化的方式在我看来也是不好的,你可能不得不到处添加Serializable。

那么解决方案是什么呢?

Java深度克隆库 克隆库是一个小型的开源(apache许可)java库,它对对象进行深度克隆。对象不必实现克隆接口。实际上,这个库可以克隆任何java对象。它可以用在缓存实现中,如果你不想修改缓存对象,或者当你想创建对象的深度副本时。

Cloner cloner=new Cloner();
XX clone = cloner.deepClone(someObjectOfTypeXX);

请登录https://github.com/kostaskougios/cloning查看

其他回答

Java 8提供了一种优雅而简洁地调用元素狗的复制构造函数或克隆方法的新方法:流、lambdas和收集器。

拷贝构造函数:

List<Dog> clonedDogs = dogs.stream().map(Dog::new).collect(toList());

表达式Dog::new被称为方法引用。它创建了一个函数对象,调用Dog上的构造函数,该构造函数以另一个Dog作为参数。

克隆方法[1]:

List<Dog> clonedDogs = dogs.stream().map(Dog::clone).collect(toList());

得到一个数组列表作为结果

或者,如果你必须得到一个数组列表(以防你以后想修改它):

ArrayList<Dog> clonedDogs = dogs.stream().map(Dog::new).collect(toCollection(ArrayList::new));

及时更新列表

如果你不需要保留狗列表的原始内容,你可以使用replaceAll方法并在适当的地方更新列表:

dogs.replaceAll(Dog::new);

所有示例都假设导入静态java.util.stream.Collectors.*;


数组列表收集器

上一个例子中的收集器可以被做成一个util方法。因为这是一件很常见的事情,我个人喜欢它是短而漂亮的。是这样的:

ArrayList<Dog> clonedDogs = dogs.stream().map(d -> d.clone()).collect(toArrayList());

public static <T> Collector<T, ?, ArrayList<T>> toArrayList() {
    return Collectors.toCollection(ArrayList::new);
}

[1] CloneNotSupportedException异常说明:

为了使这个解决方案工作,Dog的克隆方法不能声明它抛出CloneNotSupportedException。原因是map的参数不允许抛出任何受控异常。

是这样的:

    // Note: Method is public and returns Dog, not Object
    @Override
    public Dog clone() /* Note: No throws clause here */ { ...

然而,这应该不是一个大问题,因为这是最佳实践。(例如,effective Java给出了这个建议。)

感谢Gustavo注意到这一点。

一种糟糕的方法是用反思来做这件事。这种方法对我很管用。

public static <T extends Cloneable> List<T> deepCloneList(List<T> original) {
    if (original == null || original.size() < 1) {
        return new ArrayList<>();
    }

    try {
        int originalSize = original.size();
        Method cloneMethod = original.get(0).getClass().getDeclaredMethod("clone");
        List<T> clonedList = new ArrayList<>();

        // noinspection ForLoopReplaceableByForEach
        for (int i = 0; i < originalSize; i++) {
            // noinspection unchecked
            clonedList.add((T) cloneMethod.invoke(original.get(i)));
        }
        return clonedList;
    } catch (NoSuchMethodException | InvocationTargetException | IllegalAccessException e) {
        System.err.println("Couldn't clone list due to " + e.getMessage());
        return new ArrayList<>();
    }
}

基本上有三种不需要手动迭代的方法,

1使用构造函数

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>(dogs);

2使用addAll(Collection<?c)扩展;

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>();
clonedList.addAll(dogs);

3使用addAll(int index, Collection<?用int形参扩展E> c)方法

ArrayList<Dog> dogs = getDogs();
ArrayList<Dog> clonedList = new ArrayList<Dog>();
clonedList.addAll(0, dogs);

注意:如果指定的集合在操作进行时被修改,这些操作的行为将是未定义的。

List<Dog> dogs;
List<Dog> copiedDogs = dogs.stream().map(dog -> SerializationUtils.clone(dog)).Collectors.toList());

这将深度复制每个狗

一些其他用于将ArrayList复制为深度复制的替代方法

Alernative 1 -使用外部包common -lang3,方法SerializationUtils.clone():

SerializationUtils.clone()

假设我们有一个类dog,其中类的字段是可变的,并且至少有一个字段是String类型和mutable类型的对象——而不是基本数据类型(否则浅拷贝就足够了)。

浅拷贝的例子:

List<Dog> dogs = getDogs(); // We assume it returns a list of Dogs
List<Dog> clonedDogs = new ArrayList<>(dogs);

现在回到狗的深度复制。

Dog类只有可变字段。

狗类:

public class Dog implements Serializable {
    private String name;
    private int age;

    public Dog() {
        // Class with only mutable fields!
        this.name = "NO_NAME";
        this.age = -1;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    @Override
    public String toString() {
        return "Dog{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

注意,类Dog实现了Serializable!这使得可以使用方法“SerializationUtils.clone(dog)”

阅读main方法中的注释以理解结果。这表明我们已经成功地对ArrayList()进行了深度复制。看到 在“SerializationUtils.clone(dog)”下面:

public static void main(String[] args) {
    Dog dog1 = new Dog();
    dog1.setName("Buddy");
    dog1.setAge(1);

    Dog dog2 = new Dog();
    dog2.setName("Milo");
    dog2.setAge(2);

    List<Dog> dogs = new ArrayList<>(Arrays.asList(dog1,dog2));

    // Output: 'List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]'
    System.out.println("List dogs: " + dogs);

    // Let's clone and make a deep copy of the dogs' ArrayList with external package commons-lang3:
    List<Dog> clonedDogs = dogs.stream().map(dog -> SerializationUtils.clone(dog)).collect(Collectors.toList());
    // Output: 'Now list dogs are deep copied into list clonedDogs.'
    System.out.println("Now list dogs are deep copied into list clonedDogs.");

    // A change on dog1 or dog2 can not impact a deep copy.
    // Let's make a change on dog1 and dog2, and test this
    // statement.
    dog1.setName("Bella");
    dog1.setAge(3);
    dog2.setName("Molly");
    dog2.setAge(4);

    // The change is made on list dogs!
    // Output: 'List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]'
    System.out.println("List dogs after change: " + dogs);

    // There is no impact on list clonedDogs's inner objects after the deep copy.
    // The deep copy of list clonedDogs was successful!
    // If clonedDogs would be a shallow copy we would see the change on the field
    // "private String name", the change made in list dogs, when setting the names
    // Bella and Molly.
    // Output clonedDogs:
    // 'After change in list dogs, no impact/change in list clonedDogs:\n'
    // '[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]\n'
    System.out.println("After change in list dogs, no impact/change in list clonedDogs: \n" + clonedDogs);
}

输出:

List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]
Now list dogs are deep copied into list clonedDogs.
List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]
After change in list dogs, no impact/change in list clonedDogs:
[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]

备注: 因为改变列表狗后对列表克隆狗没有影响/改变, 那么ArrayList的深度复制成功!

Alernative 2 -不使用外部包:

Dog类中引入了一个新方法“clone()”,与替代方案1相比,“implements Serializable”被删除了。

clone()

狗类:

public class Dog {
    private String name;
    private int age;

    public Dog() {
        // Class with only mutable fields!
        this.name = "NO_NAME";
        this.age = -1;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    /**
     * Returns a deep copy of the Dog
     * @return new instance of {@link Dog}
     */
    public Dog clone() {
        Dog newDog = new Dog();
        newDog.setName(this.name);
        newDog.setAge(this.age);
        return newDog;
    }

    @Override
    public String toString() {
        return "Dog{" +
                "name='" + name + '\'' +
                ", age=" + age +
                '}';
    }
}

阅读下面主要方法中的评论以理解结果。这表明我们已经成功地对ArrayList()进行了深度复制。看到 下面是上下文中的“clone()”方法:

public static void main(String[] args) {
    Dog dog1 = new Dog();
    dog1.setName("Buddy");
    dog1.setAge(1);

    Dog dog2 = new Dog();
    dog2.setName("Milo");
    dog2.setAge(2);

    List<Dog> dogs = new ArrayList<>(Arrays.asList(dog1,dog2));

    // Output: 'List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]'
    System.out.println("List dogs: " + dogs);

    // Let's clone and make a deep copy of the dogs' ArrayList:
    List<Dog> clonedDogs = dogs.stream().map(dog -> dog.clone()).collect(Collectors.toList());
    // Output: 'Now list dogs are deep copied into list clonedDogs.'
    System.out.println("Now list dogs are deep copied into list clonedDogs.");

    // A change on dog1 or dog2 can not impact a deep copy.
    // Let's make a change on dog1 and dog2, and test this
    // statement.
    dog1.setName("Bella");
    dog1.setAge(3);
    dog2.setName("Molly");
    dog2.setAge(4);

    // The change is made on list dogs!
    // Output: 'List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]'
    System.out.println("List dogs after change: " + dogs);

    // There is no impact on list clonedDogs's inner objects after the deep copy.
    // The deep copy of list clonedDogs was successful!
    // If clonedDogs would be a shallow copy we would see the change on the field
    // "private String name", the change made in list dogs, when setting the names
    // Bella and Molly.
    // Output clonedDogs:
    // 'After change in list dogs, no impact/change in list clonedDogs:\n'
    // '[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]\n'
    System.out.println("After change in list dogs, no impact/change in list clonedDogs: \n" + clonedDogs);
}

输出:

List dogs: [Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]
Now list dogs are deep copied into list clonedDogs.
List dogs after change: [Dog{name='Bella', age=3}, Dog{name='Molly', age=4}]
After change in list dogs, no impact/change in list clonedDogs:
[Dog{name='Buddy', age=1}, Dog{name='Milo', age=2}]

备注: 因为改变列表狗后对列表克隆狗没有影响/改变, 那么ArrayList的深度复制成功!

注一: 方案1比方案2慢得多, 但更容易维护,因为您不需要 更新任何方法,如clone()。

注2:对于替代方案1,以下maven依赖项用于方法“SerializationUtils.clone()””:

<!-- https://mvnrepository.com/artifact/org.apache.commons/commons-lang3 -->
<dependency>
    <groupId>org.apache.commons</groupId>
    <artifactId>commons-lang3</artifactId>
    <version>3.9</version>
</dependency>

更多common-lang3版本请访问:

https://mvnrepository.com/artifact/org.apache.commons/commons-lang3