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Subject[PATCH] [RFC] dev_alloc_name() requirement of %d and the 100 names limit
Hello,

Somewhere between 2.4.10 and 2.4.12 the dev_alloc_name() function was
changed to explicitly check for a %d in the name of a to-be allocated
network interface. The idea is that this function returns a unique name
or bails out.

The recent change broke at least tinc (a userspace VPN daemon), because
it tries to alloc a tun/tap device and gives it a name, but does not require
that the %d must be present (the documentation of the tun/tap driver nor
the example code in Documentation/ make any mention of that). I don't
see why the %d would have to be mandatory, so the attached patch allows
device names without a %d to work as well.

A comment in that function says that if anyone would need more than 100
devices the algorithm should be fixed. The patch does this, allowing a
virtually unlimitted amount of names, in O(log(n)) time, n being the
number of already allocated names.

I stresstested the patched, I was able to allocate 1901 network devices
using "test%d" as the name. After that my system claimed it didn't have
any file descriptors left.

--
Met vriendelijke groet / with kind regards,
Guus Sliepen <guus@sliepen.warande.net>
--- dev.c.old Sun Oct 21 15:39:44 2001
+++ dev.c Sun Oct 21 17:45:11 2001
@@ -550,30 +550,65 @@

int dev_alloc_name(struct net_device *dev, const char *name)
{
- int i;
- char buf[32];
+ int i, j, k;
+ char buf[IFNAMSIZ];
char *p;

/*
* Verify the string as this thing may have come from
- * the user. There must be one "%d" and no other "%"
+ * the user. There may only be one "%d" and no other "%"
* characters.
*/
p = strchr(name, '%');
- if (!p || p[1] != 'd' || strchr(p+2, '%'))
+ if (!p)
+ {
+ snprintf(buf,sizeof(buf),name);
+ if (__dev_get_by_name(buf) == NULL) {
+ strcpy(dev->name, buf);
+ return 0;
+ }
+ else
+ return -ENFILE;
+ }
+
+ if (p[1] != 'd' || strchr(p+2, '%'))
return -EINVAL;

/*
- * If you need over 100 please also fix the algorithm...
+ * New algorithm that works in O(log(n)) worst case time, where
+ * n is the number of already allocated devices with the
+ * same base name.
+ *
+ * The algorithm works by trying a device number j between i and k,
+ * and changes i and k depending on whether it was allocated or not.
+ *
+ * i is the lower bound of already allocated devices,
+ * k is the upper bound of already allocated devices,
+ * k = 0 means k is infinite.
*/
- for (i = 0; i < 100; i++) {
- snprintf(buf,sizeof(buf),name,i);
+ i = j = k = 0;
+ for (;;)
+ {
+ snprintf(buf,sizeof(buf),name,j);
if (__dev_get_by_name(buf) == NULL) {
- strcpy(dev->name, buf);
- return i;
+ if (j == i || j == i + 1)
+ {
+ strcpy(dev->name, buf);
+ return j;
+ }
+ else
+ k = j;
}
+ else
+ i = j;
+
+ if (k)
+ j = (i + k) / 2;
+ else
+ j *= 2;
+ if (j == i)
+ j++;
}
- return -ENFILE; /* Over 100 of the things .. bail out! */
}

/**[unhandled content-type:application/pgp-signature]
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