问题描述
我正在尝试编写一个返回我的计算机 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/).我遇到了静态函数的 return 方面的问题.我得到的错误是 缺少返回语句.我该如何补救?
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
所有分支都必须返回一些东西,只需在末尾添加一个return 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. }
这是语法正确的 - 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:
- 异常 - 您真的只想在 System.err 上打印它们吗?
- 如果您只需要有效的地址,您是否要打印它们?
编辑在这种情况下,控制如何流动 - 正如 OP 询问最后的 return null 是否会在成功执行中否定先前的值:
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:
- enter 方法 - 新堆栈帧(1. 在代码中)
- 进入 try 块(2. 在代码中)
- 在 try 中处理指令(3. 在代码中)
- return语句:停止执行block,将值返回到上一个栈帧(代码中的4.)
- (现在不是这样,但是如果有一个 finally 块,现在就会执行,甚至会覆盖返回的值...)
- 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):
- enter 方法 - 新堆栈帧(1. 在代码中)
- 进入 try 块(2. 在代码中)
- 在 try 中处理指令(3. 在代码中)
- 抛出异常
- 进程捕获块(日志异常,6. in code)
- (现在不是这样,但是如果有一个 finally 块,现在就会执行,甚至会覆盖返回的值...)
- 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,在成功的情况下,return null; 语句,即使它在真正的返回"之后,也不会影响返回值.每当达到返回时,当前块的执行就会停止.(如果实际上下文中有一个,则相应的 finally 块将获得控制权).
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 块很棘手:仔细阅读它,这将是有用的知识.
The finally block is tricky though: read up on that, it will be useful knowledge.
- 进入 try 块(2. 在代码中)
- enter try block (2. in code)
- 进入 try 块(2. 在代码中)