Question

当我用来运行这个程序时我遇到问题。这是不断重复案例1 .`

import java.util.Scanner;    //importing 

public class DynamicStackByArray {   // name of class

    private static String[] a;
    private static int size;

    public DynamicStackByArray(){} // end of the null constructor

    public DynamicStackByArray(int capacity){

        a=new String[capacity];
    }// end of the  parameterized constructor


    private static String peek(){
        if(size==0)
            throw new IllegalStateException("Stack is Empty");
        else 
            return a[size-1];
    }// end of the displaying method

    private static String pop(){
        if(size==0)
            throw new IllegalStateException("Stack is Empty...");
        else{
            String o=a[--size];
            a[size]=null;
            return o;
        }
        }// end of the pop method
    private static void push(String o){
        if(size==a.length)resize();
        else
            a[size++]=o;
    }// end of the push method

    private static void resize(){
        String aa[]=a;
        a=new String[1+aa.length];
        System.arraycopy(aa, 0, a, 0, size);
        aa=null;
    }// end of the resize method

    public static void main(String[] args){
        Scanner in=new Scanner(System.in);

        System.out.println("Enter the size of dynamic stack :");
        int cap=in.nextInt();

        DynamicStackByArray d=new DynamicStackByArray(cap);

        System.out.println("Enter your choice \n1) for push\n2) for pop\n3) for peek");
        System.out.println("Enter -999 to exit");

        int choice=in.nextInt();

        while(choice!=-999){
            switch(choice){

            case 1:{
                System.out.println("Enter your desire data");
                System.out.println("Note that -999 will let u escape out of it... :)");

                String input;
                input=in.nextLine();
                String as="-999";

                if(!input.equalsIgnoreCase(as)){
                    d.push(input);

                }// end of checker
                System.out.println("Enter your choice \n1) for push\n2) for pop\n3) for peek");
                System.out.println("Enter -999 to exit");

                 choice=in.nextInt();
                break;
            }// end of case 1
            case 2:{
                System.out.println("Value "+d.peek()+" has been poped ");
                d.pop();
                System.out.println("Enter your choice \n1) for push\n2) for pop\n3) for peek");
                System.out.println("Enter -999 to exit");

                 choice=in.nextInt();
                break;
            }// end of the case 2
            case 3:{
                System.out.println("Value "+d.peek()+" has been peeked");
                System.out.println("Enter your choice \n1) for push\n2) for pop\n3) for peek");
                System.out.println("Enter -999 to exit");

                 choice=in.nextInt();
                break;
            }// end of the case 3
            default:{
                System.out.println("Sorry this is a wrong choice...");
                System.out.println("Enter your choice \n1) for push\n2) for pop\n3) for peek");
                System.out.println("Enter -999 to exit");

                 choice=in.nextInt();
                 break;
            }// end of default

            }// end of the switch statement
        }// end of the repetation of the program

        System.out.println("GOOD BYE... !!!");
    }// end of the main method
}// end of the class

`

Answer 1

你忘了询问扫描仪在你的循环内输入。你可以解决它：

int choice = 0;
while ((choice = in.nextInt()) != -999) {

    //... your loop
}

更重要的是：
在case 1:中，当您阅读input=in.nextLine();时，它将消耗用于终止＆＃34; 1＆＃34;的\n符号输入，您的String input将始终为空字符串。（只需输入断点并自行验证）
因此，您必须在案例1中添加额外的nextLine()语句：

in.nextLine(); // read `\n` symbol;
String input = in.nextLine();

Answer 2

它不是一直只执行case1。该程序运行正常。你在两种情况下都采用peek方法犯了错误，你应该在case2中使用pop方法作为你想要的。

任何身体plx可以帮助我...

2 个答案: