Question

我在命令提示符下使用-Xlint来解决基于Stack的程序，但我的java程序正在收到警告。我的代码是 StackDemo.java

import java.util.*;
public class StackDemo {

    static void showpush(Stack st, int a) {
        st.push(new Integer(a));
        System.out.println("push(" + a + ")");
        System.out.println("stack: " + st);
    }

    static void showpop(Stack st) {
        System.out.println("pop ->");
        Integer a = (Integer)st.pop();
        System.out.println(a);
        System.out.println("stack:" + st);
    }

    public static void main(String args[]) {
        Stack st = new Stack();
        System.out.println("stack:" + st);
        showpush(st, 42);
        showpush(st, 66);
        showpush(st, 99);
        showpop(st);
        showpop(st);
        showpop(st);
        showpop(st);
        try {
            showpop(st);
        }
        catch (EmptyStackException e) {
            System.out.println("empty stack");
        }
    }
}

Answer 1

将Stack更改为Stack<Integer>

import java.util.*;

public class StackDemo {

static void showpush(Stack<Integer> st, int a) {
  st.push(new Integer(a));
  System.out.println("push(" + a + ")");
  System.out.println("stack: " + st);
}

static void showpop(Stack<Integer> st) {
    System.out.println("pop ->");
    Integer a = st.pop();
    System.out.println(a);
    System.out.println("stack:" +st);
}

public static void main(String args[]) {
  Stack<Integer> st = new Stack<Integer>();
  System.out.println("stack:" +st);
  showpush(st, 42);
  showpush(st, 66);
  showpush(st, 99);
  showpop(st);
      showpop(st);
  showpop(st);
  showpop(st);
  try {
    showpop(st);
    }
   catch(EmptyStackException e) {
    System.out.println("empty stack");
    }
}

}

使用-Xlint后我得到警告我试图使用Stack运行程序？

1 个答案: