java - Java ACM 中的凯撒密码

Question

我在使用 ACM 的 Java 中使用凯撒密码时遇到问题。这是我的代码:

import acm.program.*;

public class Ceasar extends ConsoleProgram{

  public void run() {
    println("This program encodes a
    message using a Caesar cipher.");

    int shifter=readInt("Enter the number of character positions to shift: ");
    String msg=readLine("Enter a message :");
    String Solution=encodeCaesarCipher(msg,shifter);

    println("Encoded message: "+Solution);
  }

  private String encodeCaesarCipher(String str,int shift){
    String result="";

    for (int i=0;i<str.length();i++){
      char helper=str.charAt(i);
      helper=(helper+shift);

      if (helper>'Z'||helper>'z') helper =(helper-26);
      if (helper<'A'||helper<'a') helper=(helper+26);

      result= result+helper;
    }

    return result;
  }
}

当我编译时出现这些错误:

Ceasar.java:21: error: possible loss of precision
  helper=helper+shift;
                ^
  required: char
  found: int

Ceasar.java:22: error: possible loss of precision
  if (helper>'Z'||helper>'z') helper =helper-26;
                                             ^
  required: char
  found: int

Ceasar.java:23: error: possible loss of precision
  if (helper<'A'||helper<'a') helper=helper+26;
                                            ^
  required: char
  found: int
3 errors

Answer 1

您不能在 Java 中将 int 添加到 char 中，除非明确声明您同意可能的精度损失(溢出/下溢)。在 int 与 char 一起使用的每个地方添加一个 (char) 转换。