- 问题引入
上图中，赋给b海象的weight会改变a海象的weight，但x的赋值又不会改变y的赋值
Bits
要解释上图的问题，我们应该从Java的底层入手
相同的二进制编码，却因为数据类型不同，输出不同的值
变量的声明
基本类型
Java does not write anything into the reserved box when a variable is declared. In other words, there are no default values. As a result, the Java compiler prevents you from using a variable until after the box has been filled with bits using the = operator. For this reason, I have avoided showing any bits in the boxes in the figure above.
引用类型
When we declare a variable of any reference type (Walrus, Dog, Planet, array, etc.), Java allocates a box of 64 bits, no matter what type of object.
96位大于64位，这似乎有些矛盾：
在Java中：the 64 bit box contains not the data about the walrus, but instead the address of the Walrus in memory.
Gloden rule
练习
答案：B
解析：
- 在调用方法（函数）时，doStuff方法中的int x与main方法中的int x 实际上处于两个不同的scope（调用方法时会new 一个scope，并将main方法中的x变量的位都传递给doStuff方法中的x变量，即值传递），所以x = x - 5实际上只作用于doStuff方法中的x，而不是main方法中的x。
- 但对于引用类型来说，引用类型储存的是引用的地址，所以在进行值传递时传递的是对象的地址，所以doStuff方法中的int x与main方法中的walrus实际上指向相同的一个对象，这使得doStuff中执行的语句会作用于main方法中walrus指向的对象，所以反作用于main方法中的walrus
IntLists
使用递归求链表中元素的个数
/** Return the size of the list using... recursion! */
public int size() {
if (rest == null) {
return 1;
}
return 1 + this.rest.size();
}
Exercise: You might wonder why we don’t do something like if (this == null) return 0;. Why wouldn’t this work?
Answer: Think about what happens when you call size. You are calling it on an object, for example L.size(). If L were null, then you would get a NullPointer error!
不使用递归求元素个数
/** Return the size of the list using no recursion! */
public int iterativeSize() {
IntList p = this;
int totalSize = 0;
while (p != null) {
totalSize += 1;
p = p.rest;
}
return totalSize;
}
求第n个元素
