On 25/01/2012 16:35, Wolfram Sang wrote:

quoted

quoted

--- a/arch/arm/mach-mxs/mach-m28evk.c +++

b/arch/arm/mach-mxs/mach-m28evk.c @@ -258,7 +258,7 @@ static
int __init m28evk_fec_get_mac(void) * so hard-code DENX OUI
(C0:E5:4E) here. */ for (i = 0; i < 2; i++) { -		val = ocotp[i
* 4]; +		val = ocotp[i];

I can read that HW_OCOTP_CUST0 has the phisycal address
0x8002_C020h, and HW_OCOTP_CUST1 ist at 8002_C030h. So the two
registers are not consecutive and there are some reserved fields
between the two registers - this is the reason for the i * 4.

HW_OCOTP_CRYPTO0 is at 8002_C060h, so it is not true that the
address is currently read from the crypto-key. At least, this is
my interpretation...

Ah, that explains... that you probably never tested the code?

..and that I don't remember very well...

What about this in ocotp.c?

75         for (i = 0; i < OCOTP_WORD_COUNT; i++) 76
ocotp_words[i] = __raw_readl(ocotp_base + OCOTP_WORD_OFFSET + 77
i * 0x10);

You are right - and thanks for fixing it !

Stefano

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