java - 错误: incompatible types: void cannot be converted to NumPal

Question

这个问题已经有答案了:

What does "Incompatible types: void cannot be converted to ..." mean? (1 个回答)

已关闭 4 年前。

我读过很多关于这个问题的帖子，大多数似乎都是通过修复 main 方法来解决的。但是，任务的一部分是我们无法更改 NumPalDrvier。由于我不完全理解，我还能如何解决这个问题？

这是给定的主要代码:

import javax.swing.JOptionPane;
public class NumPalDriver {

  public static void main(String[] args){
    String start = JOptionPane.showInputDialog("Enter a number: ");
    System.out.println("start value ["+ start + "]");
    NumPal p = new NumPal(start);
    int ctr = 0;
    while (!p.pal() && (ctr < 10)){
      System.out.println(p.toString());
      System.out.print("new sum: ");
      System.out.println(p.getCur()+" + "+p.getRev()+ " = " +
                         (p.getCur()+p.getRev() ));
      p = p.next();
      ctr++;
    }
    System.out.println("final value: " + p.getCur());
    System.out.println("number of steps: " + ctr);
  }
}

我的代码使 main 成为可能:

public class NumPal{ //header

private long numRev; //initializes variables
private long numCur;
private String stringRev;
private String stringCur;

public NumPal(String stringCur){//constructor
this.stringCur = stringCur;
}

public boolean pal(){

   if (stringCur == stringRev){
      return true;}
   else {
     return false;}
     }

public String reverseString (String input){ //creates reverse version of string
   int length = input.length();
   String stringRev = "";
   for(int j = 0; j <= length; j++){
      stringRev = input.charAt(j) + stringRev; 
      }return stringRev;
      }

public void next(){ //where the error is
   stringRev = reverseString(stringCur);
   numRev = Long.parseLong(stringRev);
   numCur = Long.parseLong(stringCur);
   numCur = numCur + numRev;
   stringCur = Long.toString(numCur);
   }

public String getCur(){ // returns current string
   return stringCur;
   }

public String getRev(){ // returns reverse string
   return stringRev;
   }
}

如果有人能解释这个错误的含义，我们将不胜感激!

Answer 1

NumPalDriver 上的以下行

p = p.next();

期望 NumPal 类中的 next() 方法返回 NumPal 对象(或 null)。当前代码没有返回任何内容。重写它以返回 NumPal

即:

public NumPal next(){ 
    // return a NumPal, possibly following
    // existing code of this method goes here
    return new NumPal(stringCur)
}