Is length of ethernet header necessarily 14?

Code snippet from [here][1]:

void packet_handler(u_char *param, const struct pcap_pkthdr *header, const u_char *pkt_data)
{
....
/* retireve the position of the ip header */
ih = (ip_header *) (pkt_data +
14); //length of ethernet header
....

But this image doesn't say it's necessarily **14**:


[![alt text][2]][2]
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How should I do it properly?



[1]:

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[2]:

In 802.3, both the source and destination addresses are 48-bit MAC addresses. 6+6+2=14

The Wikipedia has a good picture of the frame

[WIKI][1]

IPv4 / v6 are layer 3 protocols.


[1]:

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The ethernet header is fixed width however extension protocols such as 802.1q for vlan/qos are common and effectivly extend the L2 header.

Yes, it's 14 in most cases for an end-station scenario.
Except the case when you have an 802.1Q frame, that would throw you off by another 4 bytes.
802.1Q is primarily used for VLAN tagging and QoS on router/router communication.

The preamble and start frame delimiter are mostly used by low level firmware to capture a frame.
By the time when we (application) have access to an ethernet frame, in general we don't have the preamble nor the start frame delimiter.

From what I can recall the 2 byte length of mac address was part of Ethernet I which never really gained acceptance. And the Ethernet II/802.3 that is having 6 byte addresses is the real common ethernet that we are using nowadays.

Also want to mention that the padding is 0-46, where 46 came from the minimum 64 bytes constraint on ethernet frame for collision detect (CD) purpose.
46(pad) + 14(dmac,smac,type) + 4(CRC) = 64 bytes