java.util.Stack类简介（栈）

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 StackTest {
     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");
         StackTest.t(s);
         System.out.println("------pop");
         String str = s.pop();
         System.out.println(str);
         StackTest.t(s);
         System.out.println("------peek");
         str = s.peek();
         System.out.println(str);
         StackTest.t(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 t(Stack<String> s){
         System.out.println("iterator");
         Iterator<String> it = s.iterator();
         while(it.hasNext()){
             System.out.print(it.next()+"    ");
         }
         System.out.println("\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）