转载自 java.util.Stack类简介
Stack是一个后进先出（last in first out，LIFO）的堆栈，在Vector类的基础上扩展5个方法而来
Deque（双端队列）比起Stack具有更好的完整性和一致性，应该被优先使用
E push(E item)
把项压入堆栈顶部。
E pop()
移除堆栈顶部的对象，并作为此函数的值返回该对象。
E peek()
查看堆栈顶部的对象，但不从堆栈中移除它。
boolean empty()
测试堆栈是否为空。
int search(Object o)
返回对象在堆栈中的位置，以 1 为基数。
Stack本身通过扩展Vector而来，而Vector本身是一个可增长的对象数组（ a growable array of objects）那么这个数组的哪里作为Stack的栈顶，哪里作为Stack的栈底？
答案只能从源代码中寻找，jdk1.6：
public class Stack<E> extends Vector<E> { /** * Creates an empty Stack. */ public Stack() { } /** * Pushes an item onto the top of this stack. This has exactly * the same effect as: * <blockquote><pre> * addElement(item)</pre></blockquote> * * @param item the item to be pushed onto this stack. * @return the <code>item</code> argument. * @see java.util.Vector#addElement */ public E push(E item) { addElement(item); return item; } /** * Removes the object at the top of this stack and returns that * object as the value of this function. * * @return The object at the top of this stack (the last item * of the <tt>Vector</tt> object). * @exception EmptyStackException if this stack is empty. */ public synchronized E pop() { E obj; int len = size(); obj = peek(); removeElementAt(len - 1); return obj; } /** * Looks at the object at the top of this stack without removing it * from the stack. * * @return the object at the top of this stack (the last item * of the <tt>Vector</tt> object). * @exception EmptyStackException if this stack is empty. */ public synchronized E peek() { int len = size(); if (len == 0) throw new EmptyStackException(); return elementAt(len - 1); } /** * Tests if this stack is empty. * * @return <code>true</code> if and only if this stack contains * no items; <code>false</code> otherwise. */ public boolean empty() { return size() == 0; } /** * Returns the 1-based position where an object is on this stack. * If the object <tt>o</tt> occurs as an item in this stack, this * method returns the distance from the top of the stack of the * occurrence nearest the top of the stack; the topmost item on the * stack is considered to be at distance <tt>1</tt>. The <tt>equals</tt> * method is used to compare <tt>o</tt> to the * items in this stack. * * @param o the desired object. * @return the 1-based position from the top of the stack where * the object is located; the return value <code>-1</code> * indicates that the object is not on the stack. */ public synchronized int search(Object o) { int i = lastIndexOf(o); if (i >= 0) { return size() - i; } return -1; } /** use serialVersionUID from JDK 1.0.2 for interoperability */ private static final long serialVersionUID = 1224463164541339165L;
} 通过peek()方法注释The object at the top of this stack (the last item of the Vector object，可以发现数组（Vector）的最后一位即为Stack的栈顶
pop、peek以及search方法本身进行了同步
push方法调用了父类的addElement方法
empty方法调用了父类的size方法
Vector类为线程安全类
综上，Stack类为线程安全类(多个方法调用而产生的数据不一致问题属于原子性问题的范畴)
public class Test { public static void main(String[] args) { Stack<String> s = new Stack<String>(); System.out.println("------isEmpty"); System.out.println(s.isEmpty()); System.out.println("------push"); s.push("1"); s.push("2"); s.push("3"); Test.it(s); System.out.println("------pop"); String str = s.pop(); System.out.println(str); Test.it(s); System.out.println("------peek"); str = s.peek(); System.out.println(str); Test.it(s); System.out.println("------search"); int i = s.search("2"); System.out.println(i); i = s.search("1"); System.out.println(i); i = s.search("none"); System.out.println(i); } public static void it(Stack<String> s){ System.out.print("iterator:"); Iterator<String> it = s.iterator(); while(it.hasNext()){ System.out.print(it.next()+";"); } System.out.print("\n"); }
} 结果：
------isEmpty
true
------push
iterator:1;2;3;
------pop
3 --栈顶是数组最后一个
iterator:1;2;
------peek
2 --pop取后删掉，peek只取不删
iterator:1;2;
------search
1 --以1为基数，即栈顶为1
2 --和栈顶见的距离为2-1=1
-1 --不存在于栈中
Stack并不要求其中保存数据的唯一性，当Stack中有多个相同的item时，调用search方法，只返回与查找对象equal并且离栈顶最近的item与栈顶间距离（见源码中search方法说明）
