缺少方法中的return语句错误 [英] Missing return statement error in a method

问题描述

我正在尝试编写一个返回计算机MAC地址字符串的静态方法（函数本身在这里找到： http://www.mkyong.com/java/how-to-get-mac-address-in-java/ ）。我遇到静态函数的 return 方面的问题。我得到的错误是缺少返回语句。我如何解决这个问题？

I am trying to write a static method that returns a string of my computer's MAC address (the function itself was found here: http://www.mkyong.com/java/how-to-get-mac-address-in-java/). I am having issues with the return aspect of the static function. The error I get is the missing return statement. How do I remedy this?

static String returnMacAddress(){
        InetAddress ip;
        try{
            ip = InetAddress.getLocalHost();

            NetworkInterface network = NetworkInterface.getByInetAddress(ip);
            byte[] mac = network.getHardwareAddress();

            System.out.print("Current MAC address: ");

            StringBuilder stringBuilder = new StringBuilder();
            for(int i = 0; i < mac.length; i++){
                stringBuilder.append(String.format("%02X%s", mac[i], (i < mac.length - 1) ? "-" : ""));
            }
            return stringBuilder.toString();
        }catch(UnknownHostException e){
            e.printStackTrace();
        } catch(SocketException e){
            e.printStackTrace();
        }
    }

推荐答案

全部分支必须返回一些东西，只需在最后添加返回null; ：

All branches must return something, just add a return null; at the end:

static String returnMacAddress(){             // 1.
    InetAddress ip;
    try{                                      // 2.
        ip = InetAddress.getLocalHost();      // 3. (until return stmt)

        NetworkInterface network = NetworkInterface.getByInetAddress(ip);
        byte[] mac = network.getHardwareAddress();

        System.out.print("Current MAC address: ");

        StringBuilder stringBuilder = new StringBuilder();
        for(int i = 0; i < mac.length; i++){
            stringBuilder.append(String.format("%02X%s", mac[i], (i < mac.length - 1) ? "-" : ""));
        }
        return stringBuilder.toString();       // 4.
    }catch(UnknownHostException e){            // 5.
        e.printStackTrace();                   // 6.
    } catch(SocketException e){
        e.printStackTrace();
    }
    return null;                               // 7.
}

这是语法正确 - 但你必须考虑这意味着什么语义，如果这是所需的操作：

This is syntactically correct - but you have to think about what this means semantically, and if that is the desired operation:

• 例外 - 你真的只想在System.err上打印它们吗？
  
  
  
  • 如果您只需要地址，是否要打印它们？
  
  

  编辑
  在这种情况下控制如何流动 - 因为OP询问最后返回null 是否会否定之前的值，成功执行：

  EDIT How the control flows in this case - as OP asked if the return null at the end would negate the previous value, in a successful execution:

  
  
  • 输入方法 - 新堆栈帧（代码中为1.）
    
    
    
    • 输入try block（2.代码中）
      
      
      
      • try中的处理指令（3.代码中）
      
      
      • return语句：停止执行块，该值返回到前一个堆栈帧（代码中为4.）
      
      
      • （现在不是这样，但如果有 finally 阻止，现在将执行，甚至可以覆盖返回的值...）
      
      
      • enter method - new stack frame (1. in code)
        • enter try block (2. in code)
          • process instructions in try (3. in code)
          • return statement: stop execution of block, the value is returned to the previous stack frame (4. in code)
          • (not a case now, but if there was a finally block, that would be executed now, and that could even overwrite the returned value...)

          在不成功的情况下，（例如UnknownHostException）：

          In unsuccessful case, (UnknownHostException for example):

          
          
          • 输入方法 - 新堆栈帧（1。在代码中）
            
            
            
            • 输入try block（2.代码中）
              
              
              
              • 在try中处理指令（3.代码中）
              
              
              • 抛出异常
              
              
              
              
              • 进程捕获块（日志异常，代码中的6.）
              
              
              • （现在不是一个案例，但如果最终有一个 阻止，现在将执行，甚至可以覆盖返回的值......）
              
              
              • process catch block (log exception, 6. in code)
              • (not a case now, but if there was a finally block, that would be executed now, and that could even overwrite the returned value...)

              如您所见，成功case， return null; 语句，即使它在实际返回之后，也不会影响返回的值。每当达到返回时，停止当前块的执行。 （如果在实际上下文中有一个，则相应的finally块将获得控件）。

              As you see, in successful case, the return null; statement, even though it is after the "real return", doesn't influence the returned value. Whenever a return is reached, the eecution of the current block is stopped. (and if there is one in the actual context, the appropriate finally block will get the control).

              finally 阻塞是棘手的：阅读，这将是有用的知识。

              The finally block is tricky though: read up on that, it will be useful knowledge.

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