Question

我正在努力想象为什么这段代码会返回它返回的内容。

public class IntObject
{
private int myInt;

public IntObject() { myInt = 0; }
public IntObject(int n) { myInt = n; }
public void increment() { myInt++; }
}

驱动程序文件

public class IntObjectTest
{
    public static IntObject someMethod(IntObject obj) {
        IntObject ans = obj;
        ans.increment();
        return ans;
}


public static void main(String[] args) {
IntObject x = new IntObject(2);
IntObject y = new IntObject(7);
IntObject a = y;
x = someMethod(y);
a = someMethod(x);}

在我脑海里，程序正在进行中是的 x是y + 1或a + 1，在这种情况下是8 a是x（8）+ 1.所以它是9

当程序结束我认为它们的值是

时

x = 8

y = 9

a = 9

但是正确的值都是9.我认为我在使用别名时遇到了麻烦。任何人都可以帮我解释别名以及他们如何在这段代码中工作吗？

Answer 1

那些不是别名，它们是引用。

IntObject x = new IntObject(2); // <-- 2
IntObject y = new IntObject(7); // <-- 7
IntObject a = y;                // <-- 7
x = someMethod(y);              // <-- 8
a = someMethod(x);              // <-- 9.

致电someMethod时，

IntObject ans = obj; // ans points to obj.
ans.increment(); // same as obj.increment();
return ans;

我认为你期待

IntObject ans = obj.clone(); // ans points to a copy of obj.
ans.increment();             // now it won't modify obj.
return ans;

Answer 2

你正在失去x和a的实际参考。

public static void main(String[] args) {
IntObject x = new IntObject(2); // "new" is creating a memory block. lets call it 0x10
IntObject y = new IntObject(7); // another "new" is creating a memory block. lets call it 0x20
IntObject a = y; // here, a = 0x20
x = someMethod(y); //someMethod actually takes an element and give same element. so here, x shows 0x20
a = someMethod(x); //someMethod actually takes an element and give same element. so here, x shows 0x20, then a shows 0x20
// at the and, y does not change anytime. so all objects addressing to 0x20
}

Answer 3

someMethod（y）返回y，所以x = y;

someMethod（x）返回x，所以a = x;

==＆GT;＆GT; y = x = a

别名和Java

3 个答案: