如何通过System.out.println在控制台上不显示任何内容？

Question

Here is a question about Stack on StackOverflow.

My question might seem very very vague but if you check my program which I have written then you might understand what I am trying to ask. I have implemented the stack myself. I present the user with 3 choices. Push, Pop and View the stack. When view(display) method is called then bunch of 0s show instead of nothing. We know the stack contains nothing unless we put something on it. But since my implemented stack is stack of integers using array, the display method when called shows bunch of 0s(the default values of integers in the array). How do I show nothing instead of 0s. I know I can add ASCII for whitespace character but I think it would still violate the rule of stack(Stack should be empty when there is not element, not even code for whitespace).

Here is my program:

import java.util.Scanner;
public class StackClass
  {

public static void main(String []args)
{

    Scanner input=new Scanner(System.in);
    int choice=0;
    int push;
    Stack stack=new Stack();

    do
    {
        System.out.println("Please select a stack operation:\n1. Press 1 for adding to stack\n2. Press 2 for removing elements from stack\n3. View the stack");
        choice=input.nextInt();

        switch(choice)
            {
                case 1:
                        System.out.println("Please enter the number that you want to store to stack");
                        push=input.nextInt();
                        stack.push(push);

                case 2:
                        stack.pop();
                case 3:
                        stack.display();
            }



    }
    while((choice==1)||(choice==2)||(choice==3));

}    
}
class Stack

{

    private int size;
    private int[] stackPlaces=new int[15];
    private int stackIndex;

    Stack()
    {
        this.size=0;
        this.stackIndex=0;
    }

    public void push(int push)
    {
        if(size<15)
        {
            stackPlaces[stackIndex]=push;
            size++;
            stackIndex++;
        }
        else
        {
            System.out.println("The stack is already full. Pop some elements and then try again");
        }
    }
    public void pop()
    {
        if(size==0)
        {
        System.out.println("The stack is already empty");
        }
        else
        {
        stackPlaces[stackIndex]=0;
        size--;
        stackIndex--;
        }
    }
    public void display()
    {
        System.out.println("The stack contains:");
        for(int i=0;i<stackPlaces.length-1;i++)
            {
            System.out.println(stackPlaces[i]);
            }
    }

}

Answer 1

In display() , simply change your loop to use size for the loop condition, so that you display the logical number of elements:

for (int i=0;i < size; i++)
{
    System.out.println(stackPlaces[i]);
}

Note that your existing loop was only showing 14 of the 15 values, too...

Answer 2

You initialize an array of int-s, of size 15. The int datatype defaults to 0 (as opposed to its wrapper class Integer which defaults to null), so what you are really doing is creating an int array with 15 0's. So, when you loop through the array and print its content, you will get, well, 15 0's.

The solution is, as implied by others, exchange the loop limitation to the size of the stack (the number of elements actually added), rather than the size of the array.

Answer 3

代替for(int i=0;i<stackPlaces.length-1;i++) ，执行for(int i=0;i<stackIndex;i++)