Pablo Neira Ayuso <[email protected]> wrote:
> +#define NET_DEVICE_PATH_STACK_MAX 5
> +
> +struct net_device_path_stack {
> + int num_paths;
> + struct net_device_path path[NET_DEVICE_PATH_STACK_MAX];
> +};
[..]
> +int dev_fill_forward_path(const struct net_device *dev, const u8 *daddr,
> + struct net_device_path_stack *stack)
> +{
> + const struct net_device *last_dev;
> + struct net_device_path_ctx ctx = {
> + .dev = dev,
> + .daddr = daddr,
> + };
> + struct net_device_path *path;
> + int ret = 0, k;
> +
> + stack->num_paths = 0;
> + while (ctx.dev && ctx.dev->netdev_ops->ndo_fill_forward_path) {
> + last_dev = ctx.dev;
> + k = stack->num_paths++;
> + if (WARN_ON_ONCE(k >= NET_DEVICE_PATH_STACK_MAX))
> + return -1;
This guarantees k < NET_DEVICE_PATH_STACK_MAX, so we can fill
entire path[].
> + path = &stack->path[k];
> + memset(path, 0, sizeof(struct net_device_path));
> +
> + ret = ctx.dev->netdev_ops->ndo_fill_forward_path(&ctx, path);
> + if (ret < 0)
> + return -1;
> +
> + if (WARN_ON_ONCE(last_dev == ctx.dev))
> + return -1;
> + }
... but this means that stack->num_paths == NET_DEVICE_PATH_STACK_MAX
is possible, with k being last element.
> + path = &stack->path[stack->num_paths++];
... so this may result in a off by one?
