Question

我已经编写了这段代码来实现Wake On Lan：

InetAddress address = InetAddress.getByAddress(
        new byte[]{(byte)0xff,(byte)0xff,(byte)0xff,(byte)0xff});
DatagramPacket packet = new DatagramPacket(bytes, bytes.length, address, 7);
socket.send(packet);

但我明白了：

java.io.IOException: Permission denied
   at gnu.java.nio.VMChannel.send(Native Method)
   at gnu.java.nio.VMChannel.send(VMChannel.java:308)
   at gnu.java.net.PlainDatagramSocketImpl.send(PlainDatagramSocketImpl.java:261)
   at java.net.DatagramSocket.send(DatagramSocket.java:659)
   at com.MyApp.wakeonlan(MyApp.java:685)

问题似乎是广播地址，因为如果我更改为任何其他地址，则不会出现此错误。我怎么能避免它呢？我正在使用linux busybox，使用JamVM和GNU Classpath。

Answer 1

现在已经很晚了，但这里是类路径0.99的补丁：

diff -aur a/native/jni/java-nio/gnu_java_nio_VMChannel.c b/native/jni/java-nio/gnu_java_nio_VMChannel.c
--- a/native/jni/java-nio/gnu_java_nio_VMChannel.c  2007-12-28 18:49:56.000000000 +0100
+++ b/native/jni/java-nio/gnu_java_nio_VMChannel.c  2012-08-14 08:49:38.000000000 +0200
@@ -868,6 +868,17 @@

   elems = (*env)->GetByteArrayElements (env, addr, NULL);

+  int n = 1;
+
+  if (*((uint32_t *) elems) == 0xFFFFFFFF)
+  {
+    if (setsockopt(fd, SOL_SOCKET, SO_BROADCAST, (char *) &n, sizeof(n)) == -1)
+    {
+      JCL_ThrowException (env, IO_EXCEPTION, "unable to set broadcast option");
+      return -1;
+    }
+  }
+
   sockaddr.sin_family = AF_INET;
   sockaddr.sin_addr.s_addr = *((uint32_t *) elems);
   sockaddr.sin_port = htons (port);

这解决了这个问题。

java.io.IOException：使用Wake on Lan拒绝权限

1 个答案: