当程序执行第4行(以“for(枚举…”开头),它立即跳转到异常捕获部分.异常没有消息,对于字符串,ex.getMessage()返回null.异常编号是830034796568(十进制).

这是代码.如上所述,一年前这种方法运作良好.

谢谢你的帮助！

public String getLocalIpAddress() {
    String address= null;
    try {  
        for (Enumeration<NetworkInterface> en = NetworkInterface.getNetworkInterfaces(); en.hasMoreElements();) {
            NetworkInterface intf = en.nextElement();
            for (Enumeration<InetAddress> enumIpAddr = intf.getInetAddresses(); enumIpAddr.hasMoreElements();) {
                InetAddress inetAddress = enumIpAddr.nextElement();
                address = new String(inetAddress.getHostAddress().toString());
                if (!inetAddress.isLoopbackAddress() && address.length() < 18) {
                    return inetAddress.getHostAddress().toString();
                }
            }
        }
    } catch (SocketException ex) {
        String msg = ex.getMessage();
        //do nothing
    }
    return null;
}