Java靜態方法參數

爲什麼下面的代碼返回100 100 1 1 1而不是100 1 1 1 1？Java靜態方法參數

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() + " "); 
} 
}

爲什麼它似乎通過酒店的價值doStuff（）？

來源

2010-04-28 Blitzkr1eg

http://stackoverflow.com/questions/40480/is-java-pass-by-reference – polygenelubricants 2010-04-28 16:01:12

Java傳遞值。看到鏈接的問題。（另外，'static'與它無關）。 – polygenelubricants 2010-04-28 16:02:18

這不正是你告訴它做:-)

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

正如其他人指出的（並且是很清楚in this article解釋），Java的經過值。在這種情況下，它會將參考文獻h1的副本傳遞給doStuff。在那裏，副本被一個新的參考覆蓋（然後返回並分配到h2），但原始值h1不受影響：它仍引用房間號爲100的第一個Hotel對象。

來源

2010-04-28 16:02:33