On Mon, Sep 13, 2010 at 11:02:42PM +0200, Hervé Poussineau wrote:
IEEE 802.3 standard requires Ethernet frames to be at least 64 bytes long.
If it is not the case, they will be considered as runt frames, and may be
ignored by netcard and/or OS
Signed-off-by: Hervé Poussineau <address@hidden>
---
slirp/slirp.c | 12 ++++++++----
1 files changed, 8 insertions(+), 4 deletions(-)
diff --git a/slirp/slirp.c b/slirp/slirp.c
index 82fd9b4..2e8c017 100644
--- a/slirp/slirp.c
+++ b/slirp/slirp.c
@@ -599,9 +599,12 @@ static void arp_input(Slirp *slirp, const uint8_t *pkt,
int pkt_len)
{
struct ethhdr *eh = (struct ethhdr *)pkt;
struct arphdr *ah = (struct arphdr *)(pkt + ETH_HLEN);
- uint8_t arp_reply[ETH_HLEN + sizeof(struct arphdr)];
- struct ethhdr *reh = (struct ethhdr *)arp_reply;
- struct arphdr *rah = (struct arphdr *)(arp_reply + ETH_HLEN);
+ union {
+ uint8_t data[ETH_HLEN + sizeof(struct arphdr)];
+ uint8_t payload[64]; /* Minimum Ethernet frame size */
+ } arp_reply;
+ struct ethhdr *reh = (struct ethhdr *)arp_reply.data;
+ struct arphdr *rah = (struct arphdr *)(arp_reply.data + ETH_HLEN);
Hi,
Would you mind explaning the point of the union here? Why not just
do something like:
- uint8_t arp_reply[ETH_HLEN + sizeof(struct arphdr)];
+ uint8_t arp_reply[MAX(ETH_HLEN + sizeof(struct arphdr), 64)];