Java静态方法参数

发布时间：2020-12-15 01:59:40 所属栏目：Java 来源：网络整理

导读：为什么以下代码返回100 100 1 1 1而不是100 1 1 1 1？ public class Hotel {private int roomNr;public Hotel(int roomNr) { this.roomNr = roomNr;}public int getRoomNr() { return this.roomNr;}static Hotel doStuff(Hotel hotel) { hotel = new Hotel(1

为什么以下代码返回100 100 1 1 1而不是100 1 1 1 1？

public class Hotel {
private int roomNr;

public Hotel(int roomNr) {
    this.roomNr = roomNr;
}

public int getRoomNr() {
    return this.roomNr;
}

static Hotel doStuff(Hotel hotel) {
    hotel = new Hotel(1);
    return hotel;
}

public static void main(String args[]) {
    Hotel h1 = new Hotel(100);
    System.out.print(h1.getRoomNr() + " ");
    Hotel h2 = doStuff(h1);
    System.out.print(h1.getRoomNr() + " ");
    System.out.print(h2.getRoomNr() + " ");
    h1 = doStuff(h2);
    System.out.print(h1.getRoomNr() + " ");
    System.out.print(h2.getRoomNr() + " ");
}
}

为什么它似乎将Hotel by-value传递给doStuff()？

解决方法

它完全按照你的要求去做:-)

Hotel h1 = new Hotel(100);
System.out.print(h1.getRoomNr() + " "); // 100
Hotel h2 = doStuff(h1);
System.out.print(h1.getRoomNr() + " "); // 100 - h1 is not changed,h2 is a distinct new object
System.out.print(h2.getRoomNr() + " "); // 1
h1 = doStuff(h2);
System.out.print(h1.getRoomNr() + " "); // 1 - h1 is now changed,h2 not
System.out.print(h2.getRoomNr() + " "); // 1

正如其他人所指出的那样(并且非常清楚地解释了in this article),Java通过了价值.在这种情况下,它将引用h1的副本传递给doStuff.在那里,副本被新引用覆盖(然后返回并分配给h2),但是h1的原始值不受影响：它仍然引用第一个房间号为100的Hotel对象.

（编辑：李大同）

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