我上了蝴蝶课:
public class Butterfly extends Insect {
/**
* Field to hold the list of colors that a butterfly object is.
*/
private List<String> colors;
/**
* Constructor to initialize the fields.
*
* @param species - Species of Butterfly.
*/
public Butterfly(String species, List<String> colors) {
super(species);
this.colors = colors;
}
/**
* Constructor to initialize an existing Butterfly object.
*
* @param butterfly - Butterfly object
*/
public Butterfly(Butterfly butterfly) {
this(butterfly.getSpecies(), butterfly.getColors());
}
/**
* Getter for the colors of the butterfly.
*
* @return the colors - Colors of the butterfly.
*/
public List<String> getColors() {
return colors;
}
@Override
public String toString() {
return getSpecies() + " " + colors;
}
}
还有一个给我问题的JUnit测试用例:
@Test
void butterfly_immutable() {
List<String> colors = new ArrayList<>();
Collections.addAll(colors, "orange", "black", "white");
Butterfly b1 = new Butterfly("Monarch", colors);
Butterfly b2 = new Butterfly(b1);
// Modifying the original color list should not affect b1 or b2.
colors.set(0, "pink");
// Modifying the colors returned by the getters should not affect b1 or b2
List<String> b1Colors = b1.getColors();
b1Colors.set(1, "lime");
List<String> b2Colors = b2.getColors();
b2Colors.set(1, "cyan");
assertTrue(sameColors(List.of("orange", "black", "white"), b1.getColors()));
assertTrue(sameColors(List.of("orange", "black", "white"), b2.getColors()));
}
我的问题是:如果修改颜色本身,如何防止更改Butterfly对象的颜色。我尝试使用List.of,List.copyOf,Collections.unmodifiableList,但似乎无法弄清楚。任何帮助将不胜感激。预先谢谢!
更改行
this.colors = colors;
to
this.colors = List.copyOf(colors);
这将使Butterfly.colors
字段成为传递到构造函数中的List
的不可修改的副本。
如果您希望Butterfly
可以通过其他方式进行修改,则可以在构造函数中进行可变复制,但也必须在“ getter”中进行复制。
this.colors = ArrayList<>(colors);
public List<String> getColors() {
return List.copyOf(colors);
}
(从技术上讲,ArrayList
构造函数可以被击败,但是您通常不必为此担心。)
更改Butterfly类的getColors()方法:
public List<String> getColors() {
return colors == null ? null : new ArrayList<>(colors);
}