检索Android手机的默认网关
我正在 Android 中编写一个本机应用程序，我需要在我的应用程序上获取设备的默认网关。这是我当前获取默认网关的代码。
static int get_default_gateway(char *def_gateway, int buf_size)
{
FILE* pipe;
char buffer[128];
char result[2049];
char cmd[] = "netstat -r | grep ^default | awk '{print $2}'";
pipe = popen(cmd, "r");
if (!pipe) return 1;
memset(result, 0, sizeof(result));
while(!feof(pipe)) {
memset(buffer, 0, sizeof(buffer));
if(fgets(buffer, 128, pipe) != NULL)
{
strcat(result, buffer);
}
}
pclose(pipe);
memset(def_gateway, 0, buf_size);
strncpy (def_gateway, result, buf_size );
return 0;
}
它适用于我的 LG p500，但在某些设备上它不会返回任何内容。
我的问题是：popen()可以在android上运行吗？我在某处读到它不包含在仿生中。
还有其他方法可以获取默认网关吗？我需要用C而不是java编写。
谢谢
I am coding a native application in android and I need to get the default gateway of a device on my application. Here is my current code to get the default gateway.
static int get_default_gateway(char *def_gateway, int buf_size)
{
FILE* pipe;
char buffer[128];
char result[2049];
char cmd[] = "netstat -r | grep ^default | awk '{print $2}'";
pipe = popen(cmd, "r");
if (!pipe) return 1;
memset(result, 0, sizeof(result));
while(!feof(pipe)) {
memset(buffer, 0, sizeof(buffer));
if(fgets(buffer, 128, pipe) != NULL)
{
strcat(result, buffer);
}
}
pclose(pipe);
memset(def_gateway, 0, buf_size);
strncpy (def_gateway, result, buf_size );
return 0;
}
It works on my LG p500 but on some devices it doesn't return anything.
My question is: Does popen() works on android? I read somewhere that it is not included in bionic.
And is there any other method to get the default gateway? I need it to be written in C and not java.
Thank you
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是的，也许 popen() 应该适用于任何 Android。但不幸的是 grep 和 awk - 不是。查看 /proc/net/route - Destination 等于 00000000 的行是您的默认网关。也许你也可以使用 NETLINK_ROUTE 套接字，尽管我从未使用过它并且不能说更多。
另请参阅此相关问题。
Yea, probably popen() should work on any Android. But unfortunately grep and awk - not. Take a look at /proc/net/route - line where Destination equals to 00000000 is your default gateway. Also perhaps you can use NETLINK_ROUTE socket, though I never used it and can't say more.
See also this related question.