--- 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! */
 }
 
 /**