Question

我正在使用一个名为decrypt的函数来解密我使用bufferedreader在文件中加密的字符串，但问题是当我使用它抛出NullPointerException的函数时，任何人都可以帮助我吗？提前谢谢！

这是我的功能：

public  String Decrypt (String Word,int key)
{
   String result="";
   for (int i=0;i<Word.length();i++)
    {
        result+=(char)(Word.charAt(i)-key);
    }
   return result;
}

bufferedreader代码：

 try {
       BufferedReader out=new BufferedReader(new FileReader("array.txt"));
       String line="";
       while((line=Decrypt(out.readLine(), 30)) !=null) // unknown exception
       {
         output.append("Your String is: \n"+ line);
       }
     }catch (FileNotFoundException ex) {

          } catch (IOException ex) {

          }

Answer 1

这不干净：

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我们知道当BufferedReader按照BufferedReader API到达流的末尾时，readline将返回body { background-color: beige; color: lightblue; padding: 0; margin:0; } header { background-color: lightblue; padding: 5px 0; margin: 0; } header h1 { color: cadetblue; font-family: Arial; margin: 0; padding: 5px 15px; display: inline; } .navMenu{ display: inline; margin: 0; } .navMenu .parentMenu { display: inline-block; list-style-type: none; background-color: lightgray; padding: 5px 10px; border: thin solid darkgray; border-radius: 3px; color: honeydew; position: relative; margin: 0; } .navMenu .parentMenu a{ color: azure; } .navMenu .parentMenu .sub-menu{ display: none; } .navMenu .parentMenu:hover .sub-menu{ display: block; position: absolute; list-style-type: none; margin:0; } .parentMenu:hover .sub-menu li{ border: thin solid darkgray; padding: 4%; background-color: lightgray; color: honeydew; text-align: left; white-space: nowrap; width: 100%; list-style-type: none; margin: 0; } .parentMenu .sub-menu li:hover { background-color: lightsteelblue; } .section { background-color: wheat; color: darkslategray; padding: 5px; float: left; display: inline; width: 63%; margin: 0 1% 1% 1%; border-radius: 10px; border: thin solid khaki; box-shadow: lightgray; } #about { float: right; width: 30%; margin: 1% 1% 0 0; text-align: center; } #home{ margin: 1% 0 1% 1%; } h4, h3 { color: lightseagreen; }，但是您的代码没有正确处理这个，即您的Decrypt方法将调用{对Word参数的{1}}方法是否为null。我建议你不要试图让你的代码过于简短，而是将阅读器的阅读单独分开。如，

while((line=Decrypt(out.readLine(), 30)) !=null)

另外，您需要学习并使用Java naming conventions。变量名都应以较低的字母开头，而类名以大写字母开头。了解并遵循这一点将使我们能够更好地理解您的代码，并使您能够更好地理解其他代码。

除了2：不要将你的异常捕获块留空，因为这是Java相当于你的汽车闭上眼睛。至少打印出堆栈跟踪并明确查看Exceptions Tutorial

解密时出现空指针异常

1 个答案: