java.util.Stack類簡介

Stack是一個後進先出（last in first out，LIFO）的堆棧，在Vector類的基礎上擴展5個方法而來

Deque（雙端隊列）比起Stack具有更好的完整性和一致性，應該被優先使用

E push(E item)
         把項壓入堆棧頂部。
E pop()
         移除堆棧頂部的對象，並作爲此函數的值返回該對象。
E peek()
         查看堆棧頂部的對象，但不從堆棧中移除它。
boolean empty()
         測試堆棧是否爲空。
int search(Object o)
         返回對象在堆棧中的位置，以 1 爲基數。

Stack本身通過擴展Vector而來，而Vector本身是一個可增長的對象數組（ a growable array of objects）那麼這個數組的哪裏作爲Stack的棧頂，哪裏作爲Stack的棧底？

答案只能從源代碼中尋找，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;
}

通過peek()方法註釋The object at the top of this stack (the last item of the Vector object，可以發現數組（Vector）的最後一位即爲Stack的棧頂

pop、peek以及search方法本身進行了同步

push方法調用了父類的addElement方法

empty方法調用了父類的size方法

Vector類爲線程安全類

綜上，Stack類爲線程安全類(多個方法調用而產生的數據不一致問題屬於原子性問題的範疇)

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");
    }
}

結果：

------isEmpty
true
------push
iterator:1;2;3;
------pop
3       --棧頂是數組最後一個
iterator:1;2;
------peek
2       --pop取後刪掉，peek只取不刪
iterator:1;2;
------search
1       --以1爲基數，即棧頂爲1
2       --和棧頂見的距離爲2-1=1
-1      --不存在於棧中

Stack並不要求其中保存數據的唯一性，當Stack中有多個相同的item時，調用search方法，只返回與查找對象equal並且離棧頂最近的item與棧頂間距離（見源碼中search方法說明）