One types:
for (i = 0 ...
So one should also type:
for_each_obj (obj ...
But the upstream kernel style guidelines are insane, and so we must
instead do:
for_each_obj(obj ...
Ugly, but one must choose his battles wisely.
Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>
Maybe an attacker on the system could use the infoleak in /proc to gauge
how long a wg(8) process takes to complete and determine the number of
leading zeros. This is somewhat ridiculous, but it's possible somebody
somewhere might at somepoint care in the future, so alright.
Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>
This is designed to work with a server that follows this:
struct sockaddr_un addr = {
.sun_family = AF_UNIX,
.sun_path = "/var/run/wireguard/wguserspace0.sock"
};
int fd, ret;
ssize_t len;
socklen_t socklen;
struct wgdevice *device;
fd = socket(AF_UNIX, SOCK_DGRAM, 0);
if (fd < 0)
exit(1);
if (bind(fd, (struct sockaddr *)&addr, sizeof(addr)) < 0)
exit(1);
for (;;) {
/* First we look at how big the next message is, so we know how much to
* allocate. Note on BSD you can instead use ioctl(fd, FIONREAD, &len). */
len = recv(fd, NULL, 0, MSG_PEEK | MSG_TRUNC);
if (len < 0) {
handle_error();
continue;
}
/* Next we allocate a buffer for the received data. */
device = NULL;
if (len) {
device = malloc(len);
if (!device) {
handle_error();
continue;
}
}
/* Finally we receive the data, storing too the return address. */
socklen = sizeof(addr);
len = recvfrom(fd, device, len, 0, (struct sockaddr *)&addr, (socklen_t *)&socklen);
if (len < 0) {
handle_error();
free(device);
continue;
}
if (!len) { /* If len is zero, it's a "get" request, so we send our device back. */
device = get_current_wireguard_device(&len);
sendto(fd, device, len, 0, (struct sockaddr *)&addr, socklen);
} else { /* Otherwise, we just received a wgdevice, so we should "set" and send back the return status. */
ret = set_current_wireguard_device(device);
sendto(fd, &ret, sizeof(ret), 0, (struct sockaddr *)&addr, socklen);
free(device);
}
}
Signed-off-by: Jason A. Donenfeld <Jason@zx2c4.com>
Jason A. Donenfeld